## Algebra 2 (1st Edition)

$x=2$
According to the exercise our equation is: $(x+4)(3x+2)=48\\3x^2+2x+12x+8=48\\3x^2+14x-40=0\\(x-2)(3x+20)=0$ Thus $3x+20=0$, hence $x=-\frac{20}{3}$ (but this gives negative lengths for the sides, thus it is not a good solution) or $x-2=0\\x=2$