Solve -2x^2=3x+2

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

Solve for x (complex solution)

$$x=\frac{-\sqrt{7}i-3}{4}\approx -0.75-0.661437828i$$

$$x=\frac{-3+\sqrt{7}i}{4}\approx -0.75+0.661437828i$$

Steps Using the Quadratic Formula

Subtract 3x from both sides.

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

Subtract 2 from both sides.

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

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

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

Square -3.

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

Multiply -4 times -2.

$$x=\frac{-\left(-3\right)±\sqrt{9+8\left(-2\right)}}{2\left(-2\right)}$$

Multiply 8 times -2.

$$x=\frac{-\left(-3\right)±\sqrt{9-16}}{2\left(-2\right)}$$

Add 9 to -16.

$$x=\frac{-\left(-3\right)±\sqrt{-7}}{2\left(-2\right)}$$

Take the square root of -7.

$$x=\frac{-\left(-3\right)±\sqrt{7}i}{2\left(-2\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{7}i}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{3±\sqrt{7}i}{-4}$$

Now solve the equation x=\frac{3±\sqrt{7}i}{-4} when ± is plus. Add 3 to i\sqrt{7}.

$$x=\frac{3+\sqrt{7}i}{-4}$$

Divide 3+i\sqrt{7} by -4.

$$x=\frac{-\sqrt{7}i-3}{4}$$

Now solve the equation x=\frac{3±\sqrt{7}i}{-4} when ± is minus. Subtract i\sqrt{7} from 3.

$$x=\frac{-\sqrt{7}i+3}{-4}$$

Divide 3-i\sqrt{7} by -4.

$$x=\frac{-3+\sqrt{7}i}{4}$$

The equation is now solved.

$$x=\frac{-\sqrt{7}i-3}{4} x=\frac{-3+\sqrt{7}i}{4}$$

Steps for Completing the Square

Subtract 3x from both sides.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide -3 by -2.

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

Divide 2 by -2.

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

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.

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

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

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

Add -1 to \frac{9}{16}.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=-\frac{7}{16}$$

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}.

$$\left(x+\frac{3}{4}\right)^{2}=-\frac{7}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{-\frac{7}{16}}$$

Simplify.

$$x+\frac{3}{4}=\frac{\sqrt{7}i}{4} x+\frac{3}{4}=-\frac{\sqrt{7}i}{4}$$

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

$$x=\frac{-3+\sqrt{7}i}{4} x=\frac{-\sqrt{7}i-3}{4}$$