Quadratic Formula
A quadratic formula is an integral part of algebra. A quadratic expression can always be factorized, but the factorization process may be difficult if the zeroes of the expression are noninteger real numbers or nonreal numbers. In such cases, we can use the quadratic formula or quadratic equation formula to determine the zeroes of the expression.
What is Quadratic Formula?
In this section, we will explore the various forms of quadratic formulas and how to use them in different case scenarios. Consider an arbitrary quadratic equation:
\[a{x^2} + bx + c = 0,\,\,\,a \ne 0\]
To determine the roots of this equation, we proceed as follows:
\[\begin{align}&a{x^2} + bx =  c\\&\Rightarrow \,\,\,{x^2} + \frac{b}{a}x =  \frac{c}{a}\end{align}\]
Now, we express the left hand side as a perfect square, by introducing a new term on both sides:
\[{x^2} + \frac{b}{a}x + \underbrace {{{\left( {\frac{b}{{2a}}} \right)}^2}}_{{\rm{New}}\,{\rm{term}}} =  \frac{c}{a} + \underbrace {{{\left( {\frac{b}{{2a}}} \right)}^2}}_{{\rm{New term}}}\]
The left hand side is now a perfect square:
\[{\left( {x + \frac{b}{{2a}}} \right)^2} =  \frac{c}{a} + \frac{{{b^2}}}{{4{a^2}}} = \frac{{{b^2}  4ac}}{{4{a^2}}}\]
This is good for us, because now we can take square roots to obtain:
\[\begin{align}&x + \frac{b}{{2a}} = \pm \frac{{\sqrt {{b^2}  4ac} }}{{2a}}\\&\Rightarrow \,\,\,x = \frac{{  b \pm \sqrt {{b^2}  4ac} }}{{2a}}\end{align}\]
Thus, by completing the squares, we were able to isolate \(x\) and obtain the two roots of the equation. Let us apply this formula to a few examples.

Example 1: Using the quadratic formula, find the roots of the equation \({x^2}  7x + 6 = 0\).
Solution:
Comparing the given equation to the standard quadratic form \(a{x^2} + bx + c = 0\), we have:
\[a = 1,\,\,b =\;\;  7,\,\,c = 6\]
Now, we use the quadratic formula to find the roots:
\[\begin{align}&x = \frac{{  b \pm \sqrt {{b^2}  4ac} }}{{2a}}\\&\Rightarrow \,\,\,x = \frac{{  \left( {  7} \right) \pm \sqrt {{{\left( {  7} \right)}^2}  4\left( 1 \right)\left( 6 \right)} }}{{2\left( 1 \right)}}\\& = \frac{{7 \pm \sqrt {49  24} }}{2} = \frac{{7 \pm \sqrt {25} }}{2}\\& = \frac{{7 \pm 5}}{2} = 1,\,\,6\end{align}\]
Thus, the two roots are \(x = 1\) and \(x = 6\). This can be verified by factorization as well. The given equation can be factorized as follows:
\[\begin{align}&{x^2}  7x + 6 = 0\\&\Rightarrow \,\,\,{x^2}  6x  x + 6 = 0\\&\Rightarrow \,\,\,x\left( {x  6} \right)  1\left( {x  6} \right) = 0\\&\Rightarrow \,\,\,\left( {x  1} \right)\left( {x  6} \right) = 0\end{align}\]
Answer: The two roots are \(x = 1\) and \(x = 6\).

Example 2: Find the roots of the following equation in the variable x using quadratic formula:
\[2p\left( {1 + {x^2}} \right)  \left( {1 + {p^2}} \right)\left( {x + p} \right) = 0\]
Solution:
Writing the given equation in the standard quadratic form, we have (verify this):
\[2p{x^2}  \left( {1 + {p^2}} \right)x + p  {p^3} = 0\]
The coefficients are:
\[a = 2p,\,\,b =  \left( {1 + {p^2}} \right),\,\,c = p  {p^3}\]
Applying the quadratic formula, we have:
\[\begin{align}&\Rightarrow \,\,\,x = \frac{{\left( {1 + {p^2}} \right) \pm \sqrt {{{\left( {1 + {p^2}} \right)}^2}  4\left( {2p} \right)\left( {p  {p^3}} \right)} }}{{2\left( {2p} \right)}}\\& = \frac{{\left( {1 + {p^2}} \right) \pm \sqrt {{p^4} + 2{p^2} + 1  8p\left( {p  {p^3}} \right)} }}{{4p}}\\&= \frac{{\left( {1 + {p^2}} \right) \pm \sqrt {9{p^4}  6{p^2} + 1} }}{{4p}}\\& = \frac{{\left( {1 + {p^2}} \right) \pm \sqrt {{{\left( {3{p^2}  1} \right)}^2}} }}{{4p}}\end{align}\]
Now, we take the square root to obtain the two roots:
\[\begin{align}&x = \frac{{\left( {1 + {p^2}} \right) \pm \left( {3{p^2}  1} \right)}}{{4p}}\\& = \frac{{4{p^2}}}{{4p}},\,\,\,\,\frac{{2  2{p^2}}}{{4p}}\\& = p,\,\,\,\frac{{1  {p^2}}}{{2p}}\end{align}\]
Answer: The roots are \(p,\,\,\,\dfrac{{1  {p^2}}}{{2p}}\)