Solve 8x^2=-2x

$$8 { x }^{ 2 } =-2x$$

Solve for x

$$x=-\frac{1}{4}=-0.25$$

$$x=0$$

Steps Using Factoring

Add 2x to both sides.

$$8x^{2}+2x=0$$

Factor out x.

$$x\left(8x+2\right)=0$$

To find equation solutions, solve x=0 and 8x+2=0.

$$x=0 x=-\frac{1}{4}$$

Steps Using the Quadratic Formula

Add 2x to both sides.

$$8x^{2}+2x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 2 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Take the square root of 2^{2}.

$$x=\frac{-2±2}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-2±2}{16}$$

Now solve the equation x=\frac{-2±2}{16} when ± is plus. Add -2 to 2.

$$x=\frac{0}{16}$$

Divide 0 by 16.

$$x=0$$

Now solve the equation x=\frac{-2±2}{16} when ± is minus. Subtract 2 from -2.

$$x=-\frac{4}{16}$$

Reduce the fraction \frac{-4}{16} to lowest terms by extracting and canceling out 4.

$$x=-\frac{1}{4}$$

The equation is now solved.

$$x=0 x=-\frac{1}{4}$$

Steps for Completing the Square

Add 2x to both sides.

$$8x^{2}+2x=0$$

Divide both sides by 8.

$$\frac{8x^{2}+2x}{8}=\frac{0}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\frac{2}{8}x=\frac{0}{8}$$

Reduce the fraction \frac{2}{8} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{1}{4}x=\frac{0}{8}$$

Divide 0 by 8.

$$x^{2}+\frac{1}{4}x=0$$

Divide \frac{1}{4}, the coefficient of the x term, by 2 to get \frac{1}{8}. Then add the square of \frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{4}x+\left(\frac{1}{8}\right)^{2}=\left(\frac{1}{8}\right)^{2}$$

Square \frac{1}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=\frac{1}{64}$$

Factor x^{2}+\frac{1}{4}x+\frac{1}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{8}\right)^{2}=\frac{1}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{8}\right)^{2}}=\sqrt{\frac{1}{64}}$$

Simplify.

$$x+\frac{1}{8}=\frac{1}{8} x+\frac{1}{8}=-\frac{1}{8}$$

Subtract \frac{1}{8} from both sides of the equation.

$$x=0 x=-\frac{1}{4}$$