How do you solve ${{x}^{2}}+8x-12=0$?

Hint: We use quadratic equation solving methods to solve the problem. We use the formula of $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to find the roots of the equation. We also use the identity of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. We use the form of ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ to form the equation.

Complete step by step answer:
We know for a general equation of quadratic $a{{x}^{2}}+bx+c=0$, the value of the roots of x will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
In the given equation we have ${{x}^{2}}+8x-12=0$. The values of a, b, c is $1,8,-12$ respectively.
We put the values and get x as
$\begin{align}
  & x=\dfrac{-8\pm \sqrt{{{8}^{2}}-4\times 1\times \left( -12 \right)}}{2\times 1} \\
 & \Rightarrow x=\dfrac{-8\pm \sqrt{112}}{2}=\dfrac{-8\pm 4\sqrt{7}}{2} \\
 & \Rightarrow x=-4\pm 2\sqrt{7} \\
\end{align}$.
The roots of the equation are $x=-4\pm 2\sqrt{7}$.
Now we find the factorisation of the equation ${{x}^{2}}+8x-12=0$ using the identity of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
We first try to form one square in the form of ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.
So, ${{x}^{2}}+8x-12={{\left( x \right)}^{2}}+2\times x\times 4+{{\left( 4 \right)}^{2}}-12-{{\left( 4 \right)}^{2}}={{\left( x+4 \right)}^{2}}-28$
We take 28 as ${{\left( \sqrt{28} \right)}^{2}}={{\left( 2\sqrt{7} \right)}^{2}}$.
Therefore, we get
\[\begin{align}
  & {{x}^{2}}+8x-12=0 \\
 & \Rightarrow {{\left( x+4 \right)}^{2}}-{{\left( 2\sqrt{7} \right)}^{2}}=0 \\
 & \Rightarrow \left( x+4+2\sqrt{7} \right)\left( x+4-2\sqrt{7} \right)=0 \\
\end{align}\]
Therefore, \[\left( x+4+2\sqrt{7} \right)\left( x+4-2\sqrt{7} \right)=0\] has multiplication of two polynomials giving a value of 0. This means at least one of them has to be 0.
We get the values of x as either \[\left( x+4+2\sqrt{7} \right)=0\] or \[\left( x+4-2\sqrt{7} \right)=0\].
This gives $x=-4+2\sqrt{7},-4-2\sqrt{7}$.

The roots of the equation are $x=-4\pm 2\sqrt{7}$.

Note: We find the value of x for which the function $f\left( x \right)={{x}^{2}}+8x-12=0$. We can verify the function. This means for $x=a$, if $f\left( a \right)=0$ then $\left( x-a \right)$ is a root of $f\left( x \right)$. We can also do the same process for \[\left( x+4+2\sqrt{7} \right)\] and \[\left( x+4-2\sqrt{7} \right)\].