Find the value of x if ${{x}^{2}}-7x=-12$

Hint: To find this quadratic equation we must know that If $a{{x}^{2}}+bx+c=0$ then $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. By substituting the values from the equation in the formula of quadratic equation and by simplifying it we get the answer.

Complete step-by-step answer:
The equation given in the question is in the form of a quadratic equation. i.e. ${{x}^{2}}-7x=-12$.
By adding 12 on both the sides, we get –
${{x}^{2}}-7x+12=0$
According to quadratic equation –
If $a{{x}^{2}}+bx+c=0$ then $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
Here, we have ${{x}^{2}}-7x+12=0$
Where,
$\begin{align}
  & a=1 \\
 & b=-7 \\
 & c=12 \\
\end{align}$
Then , $x=\dfrac{-\left( -7 \right)\pm \sqrt{{{\left( -7 \right)}^{2}}-4\left( 1 \right)\left( 12 \right)}}{2\left( 1 \right)}$
$\begin{align}
  & =\dfrac{7\pm \sqrt{49-48}}{2} \\
 & =\dfrac{7\pm \sqrt{1}}{2} \\
\end{align}$
We know that $\sqrt{1}=1$
So, $x=\dfrac{7\pm 1}{2}$
$\begin{align}
  & x=\dfrac{7+1}{2}\text{ or }x=\dfrac{7-1}{2} \\
 & x=\dfrac{8}{2}\text{ or }x=\dfrac{6}{2} \\
\end{align}$
By cancelling the common factors from numerator and denominator, we get –
$x=4\text{ or }x=3$
Hence, the value of ${{x}^{2}}-7x=-12$ is $3$ or $4$.

Note: We can also find this question in another way
The given equation is ${{x}^{2}}-7x=-12$.
By adding 12 on both sides, we get –
${{x}^{2}}-7x+12=0$
Now, we will split the middle term to factorize the equation –
For this first we need to multiply the coefficient of first and last term i.e. $1\times 12=12$
Then we will find the 2 factors of 12, such that by adding or subtracting it we must get the middle coefficient which is -7.
Let the two factors of 12 be -4 and -3 so that we can add them to get -7
By adding $-4+\left( -3 \right)=-7$
So we get –
${{x}^{2}}-3x-4x+12=0$
By factoring the above equation, we get –
$x\left( x-3 \right)-4\left( x-3 \right)=0$
Where,
$\left( x-3 \right)=0\text{ or }\left( x-4 \right)=0$
$\therefore x=3\text{ or }x=4$