# How do you solve x^2-16=0?

By using the identity ${a}^{2} - {b}^{2} = \left(a - b\right) \cdot \left(a + b\right)$ we have that
${x}^{2} - 16 = 0 \implies {x}^{2} - {4}^{2} = 0 \implies \left(x - 4\right) \left(x + 4\right) = 0$
Hence the solutions are ${x}_{1} = - 4 , {x}_{2} = 4$