Solve 12x^2+9=-6x

$$12 { x }^{ 2 } +9=-6x$$

Solve for x (complex solution)

$$x=\frac{-1+\sqrt{11}i}{4}\approx -0.25+0.829156198i$$

$$x=\frac{-\sqrt{11}i-1}{4}\approx -0.25-0.829156198i$$

Steps Using the Quadratic Formula

Add 6x to both sides.

$$12x^{2}+9+6x=0$$

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.

$$12x^{2}+6x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, 6 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 12\times 9}}{2\times 12}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 12\times 9}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-6±\sqrt{36-48\times 9}}{2\times 12}$$

Multiply -48 times 9.

$$x=\frac{-6±\sqrt{36-432}}{2\times 12}$$

Add 36 to -432.

$$x=\frac{-6±\sqrt{-396}}{2\times 12}$$

Take the square root of -396.

$$x=\frac{-6±6\sqrt{11}i}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{-6±6\sqrt{11}i}{24}$$

Now solve the equation x=\frac{-6±6\sqrt{11}i}{24} when ± is plus. Add -6 to 6i\sqrt{11}.

$$x=\frac{-6+6\sqrt{11}i}{24}$$

Divide -6+6i\sqrt{11} by 24.

$$x=\frac{-1+\sqrt{11}i}{4}$$

Now solve the equation x=\frac{-6±6\sqrt{11}i}{24} when ± is minus. Subtract 6i\sqrt{11} from -6.

$$x=\frac{-6\sqrt{11}i-6}{24}$$

Divide -6-6i\sqrt{11} by 24.

$$x=\frac{-\sqrt{11}i-1}{4}$$

The equation is now solved.

$$x=\frac{-1+\sqrt{11}i}{4} x=\frac{-\sqrt{11}i-1}{4}$$

Steps for Completing the Square

Add 6x to both sides.

$$12x^{2}+9+6x=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$12x^{2}+6x=-9$$

Divide both sides by 12.

$$\frac{12x^{2}+6x}{12}=-\frac{9}{12}$$

Dividing by 12 undoes the multiplication by 12.

$$x^{2}+\frac{6}{12}x=-\frac{9}{12}$$

Reduce the fraction \frac{6}{12} to lowest terms by extracting and canceling out 6.

$$x^{2}+\frac{1}{2}x=-\frac{9}{12}$$

Reduce the fraction \frac{-9}{12} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{1}{2}x=-\frac{3}{4}$$

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.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=-\frac{3}{4}+\left(\frac{1}{4}\right)^{2}$$

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

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=-\frac{3}{4}+\frac{1}{16}$$

Add -\frac{3}{4} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=-\frac{11}{16}$$

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}.

$$\left(x+\frac{1}{4}\right)^{2}=-\frac{11}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{-\frac{11}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{\sqrt{11}i}{4} x+\frac{1}{4}=-\frac{\sqrt{11}i}{4}$$

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

$$x=\frac{-1+\sqrt{11}i}{4} x=\frac{-\sqrt{11}i-1}{4}$$