## Algebra 2 (1st Edition)

$x=-3+\sqrt{57}$
The area of a parallelogram is the product of the base and its corresponding height. Hence here: $x(x+6)=48\\x^2+6x-48=0\\(x+3)^2-3^2-48=0\\(x+3)^2=57\\x+3=\pm\sqrt{57}\\x=-3\pm\sqrt{57}$ But we know that $x$ is non-negative, thus $x=-3+\sqrt{57}$