Solve -2a^2-8a-6=a

$$- 2 a ^ { 2 } - 8 a - 6 = a$$

Solve for a

$$a=\frac{\sqrt{33}-9}{4}\approx -0.813859338$$

$$a=\frac{-\sqrt{33}-9}{4}\approx -3.686140662$$

Steps Using the Quadratic Formula

Subtract a from both sides.

$$-2a^{2}-8a-6-a=0$$

Combine -8a and -a to get -9a.

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

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

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

Square -9.

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

Multiply -4 times -2.

$$a=\frac{-\left(-9\right)±\sqrt{81+8\left(-6\right)}}{2\left(-2\right)}$$

Multiply 8 times -6.

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

Add 81 to -48.

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

The opposite of -9 is 9.

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

Multiply 2 times -2.

$$a=\frac{9±\sqrt{33}}{-4}$$

Now solve the equation a=\frac{9±\sqrt{33}}{-4} when ± is plus. Add 9 to \sqrt{33}.

$$a=\frac{\sqrt{33}+9}{-4}$$

Divide 9+\sqrt{33} by -4.

$$a=\frac{-\sqrt{33}-9}{4}$$

Now solve the equation a=\frac{9±\sqrt{33}}{-4} when ± is minus. Subtract \sqrt{33} from 9.

$$a=\frac{9-\sqrt{33}}{-4}$$

Divide 9-\sqrt{33} by -4.

$$a=\frac{\sqrt{33}-9}{4}$$

The equation is now solved.

$$a=\frac{-\sqrt{33}-9}{4} a=\frac{\sqrt{33}-9}{4}$$

Steps for Completing the Square

Subtract a from both sides.

$$-2a^{2}-8a-6-a=0$$

Combine -8a and -a to get -9a.

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

Add 6 to both sides. Anything plus zero gives itself.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide -9 by -2.

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

Divide 6 by -2.

$$a^{2}+\frac{9}{2}a=-3$$

Divide \frac{9}{2}, the coefficient of the x term, by 2 to get \frac{9}{4}. Then add the square of \frac{9}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}+\frac{9}{2}a+\left(\frac{9}{4}\right)^{2}=-3+\left(\frac{9}{4}\right)^{2}$$

Square \frac{9}{4} by squaring both the numerator and the denominator of the fraction.

$$a^{2}+\frac{9}{2}a+\frac{81}{16}=-3+\frac{81}{16}$$

Add -3 to \frac{81}{16}.

$$a^{2}+\frac{9}{2}a+\frac{81}{16}=\frac{33}{16}$$

Factor a^{2}+\frac{9}{2}a+\frac{81}{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}.

$$\left(a+\frac{9}{4}\right)^{2}=\frac{33}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a+\frac{9}{4}\right)^{2}}=\sqrt{\frac{33}{16}}$$

Simplify.

$$a+\frac{9}{4}=\frac{\sqrt{33}}{4} a+\frac{9}{4}=-\frac{\sqrt{33}}{4}$$

Subtract \frac{9}{4} from both sides of the equation.

$$a=\frac{\sqrt{33}-9}{4} a=\frac{-\sqrt{33}-9}{4}$$