How do you solve ${{x}^{2}}-24=0$?

Hint: We first keep the variables in one side and factorise the quadratic equation We can also use the identity of $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$. We get multiplication of two terms which gives 0. The individual terms will be 0 and from those polynomials we get the roots.

Complete step by step answer:
We need to find the solution of the given equation ${{x}^{2}}-24=0$.
${{x}^{2}}-24=0\Rightarrow {{x}^{2}}=24$.
Taking square root both sides we get $x=\pm \sqrt{24}=\pm 2\sqrt{6}$
We can also use the identity of $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$.
We convert to the equation by taking values as $a=x;b=\sqrt{24}$.
$\begin{align}
  & {{x}^{2}}-24=0 \\
 & \Rightarrow \left( x+\sqrt{24} \right)\left( x-\sqrt{24} \right)=0 \\
\end{align}$
Multiplication of two terms gives 0. This means at least one individual term to be 0.
We get the values of x as either $x=\sqrt{24}$ or $x=-\sqrt{24}$.
This gives $x=\pm \sqrt{24}=\pm 2\sqrt{6}$.

The given quadratic equation ${{x}^{2}}-24=0$ has 2 solutions and they are $x=\pm 2\sqrt{6}$.

Note: The highest power of the variable or the degree of a polynomial decides the number of roots or the solution of that polynomial. Quadratic equations have 2 roots. Cubic polynomials have 3. It can be both real and imaginary roots.
We can also apply the quadratic equation formula to solve the equation ${{x}^{2}}-24=0$.
We know for a general equation of quadratic $a{{x}^{2}}+bx+c=0$, the value of the roots of x will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.