First we need to factor trinomial $ \color{blue}{ x^2-4 } $ and than we use factored form to solve an equation $ \color{blue}{ x^2-4 = 0} $.

We can see that both terms are perfect squares. $ x^2 = ( 1 x )^2 ~~ \text{and} ~~ 4 = 2 ^2 $

so we can use the difference of squares formula:

$$ a^2 - b^2 = (a - b)(a + b) $$ $$ x^{2}-4 = \left(x+2\right) \left(x-2\right) $$

Step 1: Set each factor to zero and solve equations.

$$ \begin{array}{ccc} \begin{aligned} x+2 &= 0 \\ x &= -2 \end{aligned} & ~ & \begin{aligned} x-2 &= 0 \\ x &= 2 \end{aligned} \end{array} $$

THE ALTERNATIVE SOLUTION:

Isolate the squared variable term:

$$ x^2 = 4 $$

Solve for x:

$$ x_1 = \sqrt { 4 } $$$$ x_2 = -\sqrt { 4 } $$

Quadratic Equation Solver