Solve 20x^2+2x-08=0

$$20 x ^ { 2 } + 2 x - 08 = 0$$

Solve for x

$$x=-\frac{1}{10}=-0.1$$

$$x=0$$

Steps Using Factoring

Multiply 0 and 8 to get 0.

$$20x^{2}+2x-0=0$$

Reorder the terms.

$$20x^{2}+2x=0$$

Factor out x.

$$x\left(20x+2\right)=0$$

To find equation solutions, solve x=0 and 20x+2=0.

$$x=0 x=-\frac{1}{10}$$

Steps Using the Quadratic Formula

Multiply 0 and 8 to get 0.

$$20x^{2}+2x-0=0$$

Reorder the terms.

$$20x^{2}+2x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 20 for a, 2 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}}}{2\times 20}$$

Take the square root of 2^{2}.

$$x=\frac{-2±2}{2\times 20}$$

Multiply 2 times 20.

$$x=\frac{-2±2}{40}$$

Now solve the equation x=\frac{-2±2}{40} when ± is plus. Add -2 to 2.

$$x=\frac{0}{40}$$

Divide 0 by 40.

$$x=0$$

Now solve the equation x=\frac{-2±2}{40} when ± is minus. Subtract 2 from -2.

$$x=-\frac{4}{40}$$

Reduce the fraction \frac{-4}{40} to lowest terms by extracting and canceling out 4.

$$x=-\frac{1}{10}$$

The equation is now solved.

$$x=0 x=-\frac{1}{10}$$

Steps for Completing the Square

Multiply 0 and 8 to get 0.

$$20x^{2}+2x-0=0$$

Add 0 to both sides.

$$20x^{2}+2x=0+0$$

Add 0 and 0 to get 0.

$$20x^{2}+2x=0$$

Divide both sides by 20.

$$\frac{20x^{2}+2x}{20}=\frac{0}{20}$$

Dividing by 20 undoes the multiplication by 20.

$$x^{2}+\frac{2}{20}x=\frac{0}{20}$$

Reduce the fraction \frac{2}{20} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{1}{10}x=\frac{0}{20}$$

Divide 0 by 20.

$$x^{2}+\frac{1}{10}x=0$$

Divide \frac{1}{10}, the coefficient of the x term, by 2 to get \frac{1}{20}. Then add the square of \frac{1}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{10}x+\left(\frac{1}{20}\right)^{2}=\left(\frac{1}{20}\right)^{2}$$

Square \frac{1}{20} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{10}x+\frac{1}{400}=\frac{1}{400}$$

Factor x^{2}+\frac{1}{10}x+\frac{1}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{20}\right)^{2}=\frac{1}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{20}\right)^{2}}=\sqrt{\frac{1}{400}}$$

Simplify.

$$x+\frac{1}{20}=\frac{1}{20} x+\frac{1}{20}=-\frac{1}{20}$$

Subtract \frac{1}{20} from both sides of the equation.

$$x=0 x=-\frac{1}{10}$$