Answer
The quadratic formula is $x=\dfrac{-b\pm \sqrt{b^2-4bc}}{2a}$, which can be derived as shown below.
Work Step by Step
The given quadratic equation $ax^2+bx+c=0$ can give quadratic formula as follows:
Step 1: Divide throughout by $a$.
$\Rightarrow x^2+\dfrac{b}{a}x+\dfrac{c}{a}=0$
Step 2: Add as well as subtract $\left(\dfrac{\text{Coefficient of x}}{2}\right)^2=\left(\dfrac{\frac{b}{a}}{2}\right)^2=\left(\dfrac{b}{2a}\right)^2$.
$\Rightarrow x^2+\dfrac{b}{a}x+\left(\dfrac{b}{2a}\right)^2-\left(\dfrac{b}{2a}\right)^2 +\dfrac{c}{a}=0$
Step 3: Multiply and divide the term with $x$ by 2.
$\Rightarrow x^2+2\cdot\dfrac{b}{2a}\cdot x+\left(\dfrac{b}{2a}\right)^2-\left(\dfrac{b}{2a}\right)^2 +\dfrac{c}{a}=0$
Step 4: Compress the perfect square as follows:
$\left(x+\dfrac{b}{2a}\right)^2-\left(\dfrac{b}{2a}\right)^2 +\dfrac{c}{a}=0$
Step 5: Now, solve for $x$ as follows:
$\Rightarrow \left(x+\dfrac{b}{2a}\right)^2-\dfrac{b^2}{4a^2} +\dfrac{c}{a}=0$
$\Rightarrow\left(x+\dfrac{b}{2a}\right)^2=\dfrac{b^2}{4a^2} -\dfrac{c}{a}$
$\Rightarrow\left(x+\dfrac{b}{2a}\right)^2=\dfrac{b^2-4ac}{4a^2}$
$\Rightarrow x+\dfrac{b}{2a}=\pm\dfrac{\sqrt{b^2-4ac}}{2a}$
$\Rightarrow x= -\dfrac{b}{2a}\pm\dfrac{\sqrt{b^2-4ac}}{2a}$
$\Rightarrow x= \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$, which is the required quadratic formula.
