Question

Can someone solve for x

Answer 1

bearing in mind that, whenever we have an absolute value expression, is in effect a piece-wise function with a positive and a negative version of the expression, so

$\bf |x^2-4x-5|=7\implies \begin{cases} +(x^2-4x-5)=7\\\\ -(x^2-4x-5)=7 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ +(x^2-4x-5)=7\implies x^2-4x-5=7\implies x^2-4x-12=0 \\\\\\ (x-6)(x+2)=0\implies x= \begin{cases} 6\\ -2 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ -(x^2-4x-5)=7\implies x^2-4x-5=-7\implies x^2-4x+2=0 \\\\\\ (x-2)(x-2)=0\implies x = 2$