## College Algebra (6th Edition)

The solutions are $x=-4\pm\dfrac{3}{2}\sqrt{3}$
$(2x+8)^{2}=27$ Take the square root of both sides: $\sqrt{(2x+8)^{2}}=\pm\sqrt{27}$ $2x+8=\pm\sqrt{27}$ Substitute $27$ by $9\cdot3$ and simplify: $2x+8=\pm\sqrt{9\cdot3}$ $2x+8=\pm3\sqrt{3}$ Solve for $x$: $2x=-8\pm3\sqrt{3}$ $x=-\dfrac{8}{2}\pm\dfrac{3}{2}\sqrt{3}$ $x=-4\pm\dfrac{3}{2}\sqrt{3}$