Solve 2x^2+72=-12x

$$2 x ^ { 2 } + 72 = - 12 x$$

Solve for x (complex solution)

$$x=-3+3\sqrt{3}i\approx -3+5.196152423i$$

$$x=-3\sqrt{3}i-3\approx -3-5.196152423i$$

Steps Using the Quadratic Formula

Add 12x to both sides.

$$2x^{2}+72+12x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$2x^{2}+12x+72=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 12 for b, and 72 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 2\times 72}}{2\times 2}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\times 2\times 72}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-12±\sqrt{144-8\times 72}}{2\times 2}$$

Multiply -8 times 72.

$$x=\frac{-12±\sqrt{144-576}}{2\times 2}$$

Add 144 to -576.

$$x=\frac{-12±\sqrt{-432}}{2\times 2}$$

Take the square root of -432.

$$x=\frac{-12±12\sqrt{3}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-12±12\sqrt{3}i}{4}$$

Now solve the equation x=\frac{-12±12\sqrt{3}i}{4} when ± is plus. Add -12 to 12i\sqrt{3}.

$$x=\frac{-12+12\sqrt{3}i}{4}$$

Divide -12+12i\sqrt{3} by 4.

$$x=-3+3\sqrt{3}i$$

Now solve the equation x=\frac{-12±12\sqrt{3}i}{4} when ± is minus. Subtract 12i\sqrt{3} from -12.

$$x=\frac{-12\sqrt{3}i-12}{4}$$

Divide -12-12i\sqrt{3} by 4.

$$x=-3\sqrt{3}i-3$$

The equation is now solved.

$$x=-3+3\sqrt{3}i x=-3\sqrt{3}i-3$$

Steps for Completing the Square

Add 12x to both sides.

$$2x^{2}+72+12x=0$$

Subtract 72 from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}+12x=-72$$

Divide both sides by 2.

$$\frac{2x^{2}+12x}{2}=-\frac{72}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{12}{2}x=-\frac{72}{2}$$

Divide 12 by 2.

$$x^{2}+6x=-\frac{72}{2}$$

Divide -72 by 2.

$$x^{2}+6x=-36$$

Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+6x+3^{2}=-36+3^{2}$$

Square 3.

$$x^{2}+6x+9=-36+9$$

Add -36 to 9.

$$x^{2}+6x+9=-27$$

Factor x^{2}+6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+3\right)^{2}=-27$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{-27}$$

Simplify.

$$x+3=3\sqrt{3}i x+3=-3\sqrt{3}i$$

Subtract 3 from both sides of the equation.

$$x=-3+3\sqrt{3}i x=-3\sqrt{3}i-3$$