Solve -6x=2x^2

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

Solve for x

$$x=-3$$

$$x=0$$

Steps Using Factoring

Subtract 2x^{2} from both sides.

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

Factor out x.

$$x\left(-6-2x\right)=0$$

To find equation solutions, solve x=0 and -6-2x=0.

$$x=0 x=-3$$

Steps Using the Quadratic Formula

Subtract 2x^{2} from both sides.

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

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.

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

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

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

Take the square root of \left(-6\right)^{2}.

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

The opposite of -6 is 6.

$$x=\frac{6±6}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{6±6}{-4}$$

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

$$x=\frac{12}{-4}$$

Divide 12 by -4.

$$x=-3$$

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

$$x=\frac{0}{-4}$$

Divide 0 by -4.

$$x=0$$

The equation is now solved.

$$x=-3 x=0$$

Steps for Completing the Square

Subtract 2x^{2} from both sides.

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

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.

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

Divide both sides by -2.

$$\frac{-2x^{2}-6x}{-2}=\frac{0}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\left(-\frac{6}{-2}\right)x=\frac{0}{-2}$$

Divide -6 by -2.

$$x^{2}+3x=\frac{0}{-2}$$

Divide 0 by -2.

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

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.

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

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

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

$$x=0 x=-3$$