Solve 3=-2x^2-6x

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

Solve for x

$$x=\frac{\sqrt{3}-3}{2}\approx -0.633974596$$

$$x=\frac{-\sqrt{3}-3}{2}\approx -2.366025404$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

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

Subtract 3 from both sides.

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

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

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-2\right)\left(-3\right)}}{2\left(-2\right)}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\left(-2\right)\left(-3\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-6\right)±\sqrt{36+8\left(-3\right)}}{2\left(-2\right)}$$

Multiply 8 times -3.

$$x=\frac{-\left(-6\right)±\sqrt{36-24}}{2\left(-2\right)}$$

Add 36 to -24.

$$x=\frac{-\left(-6\right)±\sqrt{12}}{2\left(-2\right)}$$

Take the square root of 12.

$$x=\frac{-\left(-6\right)±2\sqrt{3}}{2\left(-2\right)}$$

The opposite of -6 is 6.

$$x=\frac{6±2\sqrt{3}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{6±2\sqrt{3}}{-4}$$

Now solve the equation x=\frac{6±2\sqrt{3}}{-4} when ± is plus. Add 6 to 2\sqrt{3}.

$$x=\frac{2\sqrt{3}+6}{-4}$$

Divide 6+2\sqrt{3} by -4.

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

Now solve the equation x=\frac{6±2\sqrt{3}}{-4} when ± is minus. Subtract 2\sqrt{3} from 6.

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

Divide 6-2\sqrt{3} by -4.

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

The equation is now solved.

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

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\left(-\frac{6}{-2}\right)x=\frac{3}{-2}$$

Divide -6 by -2.

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

Divide 3 by -2.

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

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

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

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=-\frac{3}{2}+\frac{9}{4}$$

Add -\frac{3}{2} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+3x+\frac{9}{4}=\frac{3}{4}$$

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

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract \frac{3}{2} from both sides of the equation.

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