## Algebra 2 (1st Edition)

$2$ ft
Our equation according to the exercise: $(12+2x)(8+2x)-12\cdot8=12\cdot8\\96+16x+24x+4x^2-96=96\\4x^2+40x-96=0\\x^2+10x-24=0\\(x+12)(x-2)=0$ Thus $x=-12$ or $x=2$, but $x$ has to positive, thus $x=2$.