Solve -6k^2-2=9k

$$- 6 k ^ { 2 } - 2 = 9 k$$

Solve for k

$$k=\frac{\sqrt{33}}{12}-\frac{3}{4}\approx -0.271286446$$

$$k=-\frac{\sqrt{33}}{12}-\frac{3}{4}\approx -1.228713554$$

Steps Using the Quadratic Formula

Subtract 9k from both sides.

$$-6k^{2}-2-9k=0$$

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.

$$-6k^{2}-9k-2=0$$

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

$$k=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\left(-6\right)\left(-2\right)}}{2\left(-6\right)}$$

Square -9.

$$k=\frac{-\left(-9\right)±\sqrt{81-4\left(-6\right)\left(-2\right)}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$k=\frac{-\left(-9\right)±\sqrt{81+24\left(-2\right)}}{2\left(-6\right)}$$

Multiply 24 times -2.

$$k=\frac{-\left(-9\right)±\sqrt{81-48}}{2\left(-6\right)}$$

Add 81 to -48.

$$k=\frac{-\left(-9\right)±\sqrt{33}}{2\left(-6\right)}$$

The opposite of -9 is 9.

$$k=\frac{9±\sqrt{33}}{2\left(-6\right)}$$

Multiply 2 times -6.

$$k=\frac{9±\sqrt{33}}{-12}$$

Now solve the equation k=\frac{9±\sqrt{33}}{-12} when ± is plus. Add 9 to \sqrt{33}.

$$k=\frac{\sqrt{33}+9}{-12}$$

Divide 9+\sqrt{33} by -12.

$$k=-\frac{\sqrt{33}}{12}-\frac{3}{4}$$

Now solve the equation k=\frac{9±\sqrt{33}}{-12} when ± is minus. Subtract \sqrt{33} from 9.

$$k=\frac{9-\sqrt{33}}{-12}$$

Divide 9-\sqrt{33} by -12.

$$k=\frac{\sqrt{33}}{12}-\frac{3}{4}$$

The equation is now solved.

$$k=-\frac{\sqrt{33}}{12}-\frac{3}{4} k=\frac{\sqrt{33}}{12}-\frac{3}{4}$$

Steps for Completing the Square

Subtract 9k from both sides.

$$-6k^{2}-2-9k=0$$

Add 2 to both sides. Anything plus zero gives itself.

$$-6k^{2}-9k=2$$

Divide both sides by -6.

$$\frac{-6k^{2}-9k}{-6}=\frac{2}{-6}$$

Dividing by -6 undoes the multiplication by -6.

$$k^{2}+\left(-\frac{9}{-6}\right)k=\frac{2}{-6}$$

Reduce the fraction \frac{-9}{-6} to lowest terms by extracting and canceling out 3.

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

Reduce the fraction \frac{2}{-6} to lowest terms by extracting and canceling out 2.

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

Divide \frac{3}{2}, the coefficient of the x term, by 2 to get \frac{3}{4}. Then add the square of \frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$k^{2}+\frac{3}{2}k+\frac{9}{16}=-\frac{1}{3}+\frac{9}{16}$$

Add -\frac{1}{3} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$k^{2}+\frac{3}{2}k+\frac{9}{16}=\frac{11}{48}$$

Factor k^{2}+\frac{3}{2}k+\frac{9}{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}.

$$\left(k+\frac{3}{4}\right)^{2}=\frac{11}{48}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k+\frac{3}{4}\right)^{2}}=\sqrt{\frac{11}{48}}$$

Simplify.

$$k+\frac{3}{4}=\frac{\sqrt{33}}{12} k+\frac{3}{4}=-\frac{\sqrt{33}}{12}$$

Subtract \frac{3}{4} from both sides of the equation.

$$k=\frac{\sqrt{33}}{12}-\frac{3}{4} k=-\frac{\sqrt{33}}{12}-\frac{3}{4}$$