Solve 3x^2+12x+81=15

$$3 x ^ { 2 } + 12 x + 81 = 15$$

Solve for x (complex solution)

$$x=-2+3\sqrt{2}i\approx -2+4.242640687i$$

$$x=-3\sqrt{2}i-2\approx -2-4.242640687i$$

Steps Using the Quadratic Formula

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.

$$3x^{2}+12x+81=15$$

Subtract 15 from both sides of the equation.

$$3x^{2}+12x+81-15=15-15$$

Subtracting 15 from itself leaves 0.

$$3x^{2}+12x+81-15=0$$

Subtract 15 from 81.

$$3x^{2}+12x+66=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 12 for b, and 66 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 3\times 66}}{2\times 3}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\times 3\times 66}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-12±\sqrt{144-12\times 66}}{2\times 3}$$

Multiply -12 times 66.

$$x=\frac{-12±\sqrt{144-792}}{2\times 3}$$

Add 144 to -792.

$$x=\frac{-12±\sqrt{-648}}{2\times 3}$$

Take the square root of -648.

$$x=\frac{-12±18\sqrt{2}i}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-12±18\sqrt{2}i}{6}$$

Now solve the equation x=\frac{-12±18\sqrt{2}i}{6} when ± is plus. Add -12 to 18i\sqrt{2}.

$$x=\frac{-12+18\sqrt{2}i}{6}$$

Divide -12+18i\sqrt{2} by 6.

$$x=-2+3\sqrt{2}i$$

Now solve the equation x=\frac{-12±18\sqrt{2}i}{6} when ± is minus. Subtract 18i\sqrt{2} from -12.

$$x=\frac{-18\sqrt{2}i-12}{6}$$

Divide -12-18i\sqrt{2} by 6.

$$x=-3\sqrt{2}i-2$$

The equation is now solved.

$$x=-2+3\sqrt{2}i x=-3\sqrt{2}i-2$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$3x^{2}+12x+81=15$$

Subtract 81 from both sides of the equation.

$$3x^{2}+12x+81-81=15-81$$

Subtracting 81 from itself leaves 0.

$$3x^{2}+12x=15-81$$

Subtract 81 from 15.

$$3x^{2}+12x=-66$$

Divide both sides by 3.

$$\frac{3x^{2}+12x}{3}=-\frac{66}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{12}{3}x=-\frac{66}{3}$$

Divide 12 by 3.

$$x^{2}+4x=-\frac{66}{3}$$

Divide -66 by 3.

$$x^{2}+4x=-22$$

Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+4x+2^{2}=-22+2^{2}$$

Square 2.

$$x^{2}+4x+4=-22+4$$

Add -22 to 4.

$$x^{2}+4x+4=-18$$

Factor x^{2}+4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+2\right)^{2}=-18$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{-18}$$

Simplify.

$$x+2=3\sqrt{2}i x+2=-3\sqrt{2}i$$

Subtract 2 from both sides of the equation.

$$x=-2+3\sqrt{2}i x=-3\sqrt{2}i-2$$