Solve 2x^2-2x+2+5x=0

$$2 { x }^{ 2 } -2x+2+5x=0$$

Solve for x (complex solution)

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

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

Steps Using the Quadratic Formula

Combine -2x and 5x to get 3x.

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

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}.

$$x=\frac{-3±\sqrt{3^{2}-4\times 2\times 2}}{2\times 2}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 2\times 2}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-3±\sqrt{9-8\times 2}}{2\times 2}$$

Multiply -8 times 2.

$$x=\frac{-3±\sqrt{9-16}}{2\times 2}$$

Add 9 to -16.

$$x=\frac{-3±\sqrt{-7}}{2\times 2}$$

Take the square root of -7.

$$x=\frac{-3±\sqrt{7}i}{2\times 2}$$

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}.

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

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

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

The equation is now solved.

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

Steps for Completing the Square

Combine -2x and 5x to get 3x.

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

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

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.

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