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    If all methods to solve quadratic equations are included in your syllabus, then these words “solve quadratic equations by factoring method” are not new to you. But it’s not necessary that every student is efficient in solving factorization. So, if you are also escaping from factorization and you are finding the factoring quadratics calculator then this explanation is for you. Before starting the explanation of the factoring method to solve quadratic equations we know about the factorization.


    In mathematics any number which is exactly divisible by the other number then the other number will be the factor of that number. For instance, 12 is exactly divisible by 1, 2, 3, 4, 6, & 12 then all these numbers are the factors of 12. So, any expression which is expressed in a form of product of two or more factors is known as factorization.


    The second-degree polynomial equation with its standard form as ax2 + bx + c = 0 is known as Quadratic equation and the formula for Quadratic equation is as:

    (x = \dfrac{ -b \pm \sqrt{b^2 – 4ac}}{ 2a })

    To solve quadratic equations is easy and by using the online Quadratic formula calculator this task becomes very easy with great accuracy of the result. But when the quadratic equations are solved by the factoring method then it is necessary to get know how about the process of factoring to become the best Quadratic equation solver. Let’s understand the process of factoring to solve quadratic equations.


    In the process of the factoring of the form of ax2 + bx + c = 0, we consider that the factors a and b of c (which is a third term) are found such that their sum (a + b) is equal to b (which is the coefficient of the middle term). For finding the factors of ax2 + bx + c = 0, we follow the following steps:

    • To find “a” the coefficient of x2, “b” the coefficient of x and “c” the constant term.
    • To find two numbers b & c such that b + c = b and bc = ac.
    • The factors of ax2 + bx + c will be (ax + b) and (x + c).

    EXAMPLE #1:

    Solve the equation 2a2 + a – 1 by factoring method.


    Here a = 2, b = 1 & c = -1.

    First, we multiply a with c = 2 × -1 = -2, now find the factors of -2 which are 1 & 2 and when we take it as 2 – 1 = 1 and 1 = a so, we may write it as 2a – a = a which is equal to b.

    2a2 + 2a – a – 1

    2a (a + 1) – 1(a + 1)

    (2a -1) (a +1).

    EXAMPLE #2:

    Solve the equation 6a2 + 11a – 10.


    Here a = 6, b = 11, and c = -10

    By the multiplication of a and c we get 6×-10 = -60, now find the factors of -60.

    1× -60, 2 × -30, 3 × -20, 5 × -12, 6 × -10, 10 × -6, 12 × -5, 15 × -4 & 60 × -1.

    From the above factors of -60 if we take

    15 – 4 = 11 and when we multiply these numbers, we will get the product of a × c = 60. So, 15a – 4a = 11a.

    6a2 + 15a – 4a – 10.

    3a (2a + 5) – 2(2a + 5).

    (3a – 2) (2a + 5).

    From the above discussion it will be clear to you how you will solve quadratic equations by factoring method. But if you have less time to submit your assignment and you need to find the quadratic function calculator then try our factoring quadratics calculator to solve quadratic equations in minutes with accurate results.

    Quadratic Equation Is Based On Squared

    “Formula for Quadratic equation” these words come to your mind when you start to solve Quadratic equations as a new topic in your study. As we know quad means double that’s why one variable in the Quadratic equation is based on squared. Basically, the word quad is a Latin word and when we solve the quadratic equation, we find it in its standard form as \(ax^2 + bx + c=0\) The most important method to solve quadratic equations is the Quadratic formula \(x = \dfrac{ -b \pm \sqrt{b^2 – 4ac}}{ 2a }\) The solution of the quadratic equation by this formula is very easy. We just put the values of a, b, & c in the formula then solve the equation according to the structure of the quadratic formula and get required solution sets. In this modern & scientific era online Quadratic formula calculator is also available to solve the quadratic equations in minutes with the accuracy of the result. Now we discuss who invented the Quadratic formula? or the history of the Quadratic equation formula.

    History Of The Formula For Quadratic Equation

    Babylonians Contributions:

    Through the Babylonian’s clay tablets as early as 2000 BC the Babylonian mathematicians used quadratic equations by geometric method to find out the area related problems of rectangles. They solved the quadratic equation only for the positive roots and the process which they used to find the values of x and y was as under:

    • First, they calculated the half value of p.

    • Secondly, they got the square of the result.

    • In the third step they subtracted the value of q.

    • In the fourth step they found the positive square root of the values.

    • Finally, they added the result of the first and 4th step to get the value of x.

    The above-mentioned steps may be written as in the form of formula:

    \(x = \dfrac{p}{2} + \sqrt{(\dfrac{p}{2})^2-q}\)

    The Babylonian mathematicians didn’t use the geometrical method as in the form of formula but if we compare the above-mentioned formula with the today formula as \(x=\dfrac{-b+\sqrt{b^2-4ac}}{2a} \) then in the today formula \(a=1, b=-p, \& \space c = q \space \) So, we can say that it became the origin of the formula for the quadratic equation.

    solve the quadratic equation by the quadratic formula \(x^2 – 6x + 8 = 0\)


    Here \(a=1,b=-6, \space \& \space c=8\)

    Now apply the quadratic formula:

    \(x = \dfrac{-b\pm\sqrt{b^2-4ac}}{ 2}\)

    \(x=\dfrac{-(-6)\pm\sqrt{(-6)^2-4(4)(8)}}{2(1)} \)

    \(x=\dfrac{6\pm\sqrt{36-32 }}{ 2}\)

    \(x=\dfrac{6\pm\sqrt{4 }}{ 2}\)

    \(x = \dfrac{ 6 \pm 2}{2} \)

    \(x = \dfrac{ 6 + 2 } { 2 } \)   OR   \( x = \dfrac{ 6 – 2}{2}\)

    \(x = \dfrac{8} { 2} \)   OR   \( x = \dfrac{4} {2} \)

    \(x = 4 \)   OR   \( x = 2\)

    Solution sets {4, 2}. Answer


    Solve the quadratic equation \(4x^2 + 12x = 7 \) by the Quadratic formula.


    First, we equation as: \(4x^2 + 12 x – 7 = 0 \)

    Now solve the equation here \(a = 4, b = 12, \space \& \space c = – 7 \)

    Apply the quadratic formula:

    \(x = \dfrac{- b \pm \sqrt{b^2 – 4ac}}{ 2a}\)

    \(x = \dfrac{ – 12 \pm \sqrt{(12)^2 – 4 (4) (-7) }}{ 2 (4)}\)

    Users Of The Geometric Method To Solve The Quadratic Equations:

    Besides Babylonian mathematicians, the mathematicians of China, Egypt, India and Greece also used the geometric method (not in the form of general formula) to solve the quadratic equations only for the positive roots.

    Contribution Of Indian Mathematician:

    In 628 AD an Indian mathematician named Brahmagupta explained in wording the following explicit form of formula to solve the quadratic equations as:

    \(x = \dfrac{\sqrt{4ac + b^2 } \space – b }{ 2a}\)

    Contributions Of Al Khawarizimi And Other Muslim Mathematicians:

    In 9th century an Islamic mathematician Muhammad ibn Musa al-Khwarizmi inspired by the concept of Brahmagupta and developed a set of formula to solve the positive roots of the quadratic equations. He also developed the completing square method and identified that the discriminant should be positive in solving quadratic equations. Another Muslim mathematician Hamid ibn Turk proved the working of the Khwarizmi that the discriminant should not be negative to solve the quadratic equations.

    The common thing in the above discussion is that all the mathematicians solved the quadratic equations for positive roots but later on Muslim mathematicians proved that the quadratic equations could be solved for negative roots as well as for irrational numbers.

    Contributions Of Simon Stevin And Rene Descartes:

    Finally in 1594 Simon Stevin by his work obtained the formula for Quadratic equation and in 1637 the formula for Quadratic equation which we use today as \(x = \dfrac{ – b \pm \sqrt{b^2 – 4ac }}{ 2a } \) was published by Rene Descartes in La Geometrie.

    So, the above complete history of the quadratic formula is telling us how the quadratic formula derived. The quadratic formula is the best Quadratic equation solver. There are other methods to solve the quadratic equations like completing the square method, factoring method and graphing method etc. but the quadratic formula is widely used by the people of different fields to solve the quadratic equations problems. Now we understand the Quadratic formula’s calculation by the following examples:

    EXAMPLE #1:

    \(x = \dfrac{ – 12 \pm \sqrt{144 + 112 }}{ 8}\)

    \(x = \dfrac{ – 12 \pm \sqrt{256 }}{ 8}\)

    \(x = \dfrac{ – 12 \pm 16 }{ 8}\)

    \(x = \dfrac{ – 12 + 16 }{ 8} \)   OR   \( x = \dfrac{ – 12 = 16 }{ 8} \)

    \(x = \dfrac{4} {8} \)   OR   \( x = \dfrac{ – 28 } {8} \)

    \(x = \dfrac{1 }{ 2} \)   OR   \( x = \dfrac{ – 7 } {2}\)

    Solution sets {0.5, – 3.5} Answer.

    From the above discussion and Quadratic formula calculation examples it will be clear to you how the Quadratic formula derived? And how you can solve the quadratic equation by Quadratic formula.

    If you are a student and in your daily hectic schedule have no time to complete your assignment and you are trying to find the quadratic function calculator to solve your Quadratic equations problems.

    So, solve your problems in less time by using our Quadratic function calculator or Quadratic formula calculator and be the best Quadratic equation solver in front of your class.

    Besides the quadratic formula calculator, you may also use our factoring quadratics calculator to solve your Quadratic equations problems by factoring method. We not only guide you in solving quadratic equations by different methods but also explain to you the theory of all solving methods of quadratic equations so, understand with us and be a champion of the quadratic equation solver.