Solve -2x^2=3x

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

Solve for x

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$x=0$$

Steps Using Factoring

Subtract 3x from both sides.

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

Factor out x.

$$x\left(-2x-3\right)=0$$

To find equation solutions, solve x=0 and -2x-3=0.

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

Steps Using the Quadratic Formula

Subtract 3x from both sides.

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

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

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

Take the square root of \left(-3\right)^{2}.

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

The opposite of -3 is 3.

$$x=\frac{3±3}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{3±3}{-4}$$

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

$$x=\frac{6}{-4}$$

Reduce the fraction \frac{6}{-4} to lowest terms by extracting and canceling out 2.

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

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

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

Divide 0 by -4.

$$x=0$$

The equation is now solved.

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

Steps for Completing the Square

Subtract 3x from both sides.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\left(-\frac{3}{-2}\right)x=\frac{0}{-2}$$

Divide -3 by -2.

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

Divide 0 by -2.

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

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.

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

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

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{9}{16}$$

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

$$\left(x+\frac{3}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$x+\frac{3}{4}=\frac{3}{4} x+\frac{3}{4}=-\frac{3}{4}$$

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

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