How do you solve the given quadratic equation ${{x}^{2}}-12x+32=0$?

Hint: We start solving the problem by factoring the given quadratic equation. We then take the common factors out by grouping the first two and next two terms. We then take the common terms from the obtained result and then equate each factor to 0 to get the roots of the given quadratic equation which is the required result for the given problem.

Complete step by step answer:
According to the problem, we are asked to solve the given quadratic equation ${{x}^{2}}-12x+32=0$.
We have given the quadratic equation ${{x}^{2}}-12x+32=0$.
Let us factorize this given quadratic equation to find the required roots.
$\Rightarrow {{x}^{2}}-8x-4x+32=0$.
$\Rightarrow \left( x\times x \right)+\left( x\times -8 \right)+\left( -4\times x \right)+\left( -4\times -8 \right)=0$.
Now, let us take the common factors out by grouping the first two terms and next terms together.
\[\Rightarrow x\times \left( x-8 \right)-4\times \left( x-8 \right)=0\].
We can see that the common factor is $\left( x-8 \right)$. So, let us take it common.
\[\Rightarrow \left( x-4 \right)\left( x-8 \right)=0\].
Now, let us equate each factor to get the required roots of the given quadratic equation.
\[\Rightarrow x-4=0\], \[x-8=0\].
\[\Rightarrow x=4\], $x=8$.
So, we have found the solution(s) of the given quadratic equation ${{x}^{2}}-12x+32=0$ as $x=4$, $x=8$.
$\therefore $ The solution(s) of the given quadratic equation ${{x}^{2}}-12x+32=0$ are $x=4$, $x=8$.

Note:
We can also solve the given problem as shown below:
We have given the quadratic equation ${{x}^{2}}-12x+32=0$ ---(1).
We know that the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ is $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. Let us use this result in equation (1).
So, the roots are $\dfrac{-\left( -12 \right)\pm \sqrt{{{\left( -12 \right)}^{2}}-4\left( 1 \right)\left( 32 \right)}}{2\left( 1 \right)}=\dfrac{12\pm \sqrt{144-128}}{2}=\dfrac{12\pm \sqrt{16}}{2}=\dfrac{12\pm 4}{2}=\dfrac{16}{2},\dfrac{8}{2}=8,4$.