## Algebra 2 (1st Edition)

Simplifying using the rules of radicals, we find: $$\left(x+2\right)^2=\left(\sqrt{9x+28}\right)^2 \\ x^2+4x+4=9x+28 \\ x=8,\:x=-3$$ -3 is not a valid solution, so: $$x=8$$