How do you solve $-3(x + 6)^2 = 18$ ?

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Answer 1 · 2015-07-30T00:32:16

$x=-6+sqrt6 i, -6-sqrt 6 i$

$-3(x+6)^2=18$

Divide both sides by $-3$ .

$(x+6)^2=18/-3$ =

$(x+6)^2=-6$

Take the square root of both sides.

$x+6=+-sqrt(-6)$

Subtract $6$ from both sides.

$x=-6+-sqrt(-6)$

Solve for $x$ .

$x=-6+sqrt6i$

$x=-6-sqrt6i$

How do you solve -3(x + 6)^2 = 18-3(x + 6)^2 = 18?