## Precalculus: Mathematics for Calculus, 7th Edition

$x=1$
$(x+2)^{2}=(x-4)^{2}$ Evaluate the powers on both sides: $x^{2}+4x+4=x^{2}-8x+16$ Since $x^{2}$ is repeated on both sides, it can be removed from the equation $4x+4=-8x+16$ Take $-8x$ to the left side and $4$ to the right side: $4x+8x=16-4$ Evaluate the operations on both sides: $12x=12$ Take $12$ to divide the right side: $x=\dfrac{12}{12}=1$ $x=1$