Solve 3x+6=-6x^2

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

Solve for x (complex solution)

$$x=\frac{-1+\sqrt{15}i}{4}\approx -0.25+0.968245837i$$

$$x=\frac{-\sqrt{15}i-1}{4}\approx -0.25-0.968245837i$$

Steps Using the Quadratic Formula

Add 6x^{2} to both sides.

$$3x+6+6x^{2}=0$$

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.

$$6x^{2}+3x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, 3 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\times 6\times 6}}{2\times 6}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 6\times 6}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-3±\sqrt{9-24\times 6}}{2\times 6}$$

Multiply -24 times 6.

$$x=\frac{-3±\sqrt{9-144}}{2\times 6}$$

Add 9 to -144.

$$x=\frac{-3±\sqrt{-135}}{2\times 6}$$

Take the square root of -135.

$$x=\frac{-3±3\sqrt{15}i}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{-3±3\sqrt{15}i}{12}$$

Now solve the equation x=\frac{-3±3\sqrt{15}i}{12} when ± is plus. Add -3 to 3i\sqrt{15}.

$$x=\frac{-3+3\sqrt{15}i}{12}$$

Divide -3+3i\sqrt{15} by 12.

$$x=\frac{-1+\sqrt{15}i}{4}$$

Now solve the equation x=\frac{-3±3\sqrt{15}i}{12} when ± is minus. Subtract 3i\sqrt{15} from -3.

$$x=\frac{-3\sqrt{15}i-3}{12}$$

Divide -3-3i\sqrt{15} by 12.

$$x=\frac{-\sqrt{15}i-1}{4}$$

The equation is now solved.

$$x=\frac{-1+\sqrt{15}i}{4} x=\frac{-\sqrt{15}i-1}{4}$$

Steps for Completing the Square

Add 6x^{2} to both sides.

$$3x+6+6x^{2}=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

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

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.

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

Divide both sides by 6.

$$\frac{6x^{2}+3x}{6}=-\frac{6}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\frac{3}{6}x=-\frac{6}{6}$$

Reduce the fraction \frac{3}{6} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{1}{2}x=-\frac{6}{6}$$

Divide -6 by 6.

$$x^{2}+\frac{1}{2}x=-1$$

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=-1+\left(\frac{1}{4}\right)^{2}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=-1+\frac{1}{16}$$

Add -1 to \frac{1}{16}.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=-\frac{15}{16}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{4}\right)^{2}=-\frac{15}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{-\frac{15}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{\sqrt{15}i}{4} x+\frac{1}{4}=-\frac{\sqrt{15}i}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

$$x=\frac{-1+\sqrt{15}i}{4} x=\frac{-\sqrt{15}i-1}{4}$$