How do you solve this by completing the square: $a{{x}^{2}}+bx+c=0$ ?

Hint:
In the above question, we have to solve the equation $a{{x}^{2}}+bx+c=0$ . This is a linear equation so we may have more than one solution for this problem. Also, the a, b, and c are variables and coefficients of ${{x}^{2}}$ , x, and 1 respectively. So let’s solve this problem.

Complete step by step solution:
The given equation is $a{{x}^{2}}+bx+c=0$ .
On dividing both the sides with a, we get
$\Rightarrow {{x}^{2}}+\dfrac{b}{a}x+\dfrac{c}{a}=0$
After subtracting the constant term from both sides of the equation, we will get
$\Rightarrow {{x}^{2}}+\dfrac{b}{a}x=-\dfrac{c}{a}$
To make the above equation a perfect square, the constant term should be ${{(\dfrac{b}{2a})}^{2}}$
After adding ${{(\dfrac{b}{2a})}^{2}}$ on both sides of the equation, we get
$\Rightarrow {{x}^{2}}+\dfrac{b}{a}x+{{(\dfrac{b}{2a})}^{2}}={{(\dfrac{b}{2a})}^{2}}-\dfrac{c}{a}$
Now, we have to rewrite the left side to make it a perfect square
$\Rightarrow {{(x+(\dfrac{b}{2a}))}^{2}}={{(\dfrac{b}{2a})}^{2}}-\dfrac{c}{a}$
Taking the square root on both the sides
$\Rightarrow x+(\dfrac{b}{2a})=\pm \sqrt{{{(\dfrac{b}{2a})}^{2}}-\dfrac{c}{a}}$
Subtracting the constant term of the left side from both the sides
$\Rightarrow x=\pm \sqrt{{{(\dfrac{b}{2a})}^{2}}-\dfrac{c}{a}}-(\dfrac{b}{2a})$
After solving the above expression we will get
$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

Therefore, after solving $a{{x}^{2}}+bx+c=0$ we get $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

Note:
In the above solution, you can see there are two answers, the first one is $\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}$ and the second one is $\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}$ . As we were solving a linear equation therefore, we got two answers. $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ is known as Shreedhara Acharya's formula.