Question

$frac{(x-4)^2}{3}-frac{(x+4)^2}{3} =5$

Answer 1

$frac{(x-4)^2}{3} - frac{(x+4)^2}{3} =5, |cdot 3\\(x-4)^2-(x+4)^2=15\\x^2-8x+16-(x^2+8x+16)=15\\-16x=15\\x=-frac{15}{16}=-0,9375$