Answer
$x = (-2 + 3\sqrt 2, -2 - 3\sqrt 2)$
Work Step by Step
$(x + 2)^2 = 18$
$x + 2 = ±\sqrt 18$ (Use Radical Property)
$x + 2 = ±3\sqrt 2$ (Simplify the radical)
$x = -2 ± 3\sqrt 2$ (Subtract 2 from both sides)
Check 1:
Let $x = -2+3\sqrt 2$
$(x+2)^2 = 18$
$((-2+3\sqrt 2)+2)^2$ could be $18$
$(-2+3\sqrt 2+2)^2$ could be $18$
$(3\sqrt 2)^2$ could be $18$
$9\times2$ could be $18$
$18=18$ True.
Check 2:
Let $x = -2+3\sqrt 2$
$(x+2)^2 = 18$
$((-2-3\sqrt 2)+2)^2$ could be $18$
$(-2-3\sqrt 2+2)^2$ could be $18$
$(-3\sqrt 2)^2$ could be $18$
$9\times2$ could be $18$
$18=18$ True.
Therefore the solution set is $(-2 + 3\sqrt 2, -2 - 3\sqrt 2)$
