How do you solve ${{x}^{2}}-4=0$?

Hint: To solve above question we will use the concept of the quadratic equation. Let us assume that $a{{x}^{2}}+bx+c=0$ as a general quadratic equation then the root of this quadratic equation is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . Here, ${{b}^{2}}-4ac$ is also known as the discriminant of the equation.

Complete step by step answer:
We can see that the given question is an equation of degree 2. So, it is generally a quadratic equation. Hence, we will use the concept of the root of the quadratic equation to solve the above equation.
We know that for a general quadratic equation $a{{x}^{2}}+bx+c=0$, its root or we can say its solution is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. Also, we will have a real root only when ${{b}^{2}}-4ac\ge 0$.
Now, we will compare the given equation ${{x}^{2}}-4=0$ with the general equation $a{{x}^{2}}+bx+c=0$ to find the coefficient a, b and c so, that by substituting their value in $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, we get the solution of given equation.
So, after comparing them we will get:
a = 1, b = 0 (because we don’t have x in ${{x}^{2}}-4=0$), and c = -4.
So, after putting value of a, b, and c in $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, we will get:
$x=\dfrac{-0\pm \sqrt{{{0}^{2}}-4\times 1\times \left( -4 \right)}}{2\times 1}$
$\Rightarrow x=\dfrac{\pm \sqrt{16}}{2}$
$\Rightarrow x=\pm 2$

Hence, x = 2, -2 is the required solution and root of ${{x}^{2}}-4=0$.

Note: We can also solve the above given equation ${{x}^{2}}-4=0$ by first factoring ${{x}^{2}}-4=0$ and then equating each of the factor to 0 to get required solution.
So, we can also write ${{x}^{2}}-4=0$ as ${{x}^{2}}-{{\left( 2 \right)}^{2}}=0$
And, we know that ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ .
Hence, we can write ${{x}^{2}}-{{\left( 2 \right)}^{2}}=\left( x-2 \right)\left( x+2 \right)=0$
Now, after equating (x - 2) and (x + 2) both to 0, we will get:
x = -2, 2.