Solve x/5=x+x^2

$$\frac { x } { 5 } = x + x ^ { 2 }$$

Solve for x

$$x=-\frac{4}{5}=-0.8$$

$$x=0$$

Steps Using Factoring

Multiply both sides of the equation by 5.

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

Subtract 5x from both sides.

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

Combine x and -5x to get -4x.

$$-4x=5x^{2}$$

Subtract 5x^{2} from both sides.

$$-4x-5x^{2}=0$$

Factor out x.

$$x\left(-4-5x\right)=0$$

To find equation solutions, solve x=0 and -4-5x=0.

$$x=0 x=-\frac{4}{5}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 5.

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

Subtract 5x from both sides.

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

Combine x and -5x to get -4x.

$$-4x=5x^{2}$$

Subtract 5x^{2} from both sides.

$$-4x-5x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-5x^{2}-4x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, -4 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}}}{2\left(-5\right)}$$

Take the square root of \left(-4\right)^{2}.

$$x=\frac{-\left(-4\right)±4}{2\left(-5\right)}$$

The opposite of -4 is 4.

$$x=\frac{4±4}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{4±4}{-10}$$

Now solve the equation x=\frac{4±4}{-10} when ± is plus. Add 4 to 4.

$$x=\frac{8}{-10}$$

Reduce the fraction \frac{8}{-10} to lowest terms by extracting and canceling out 2.

$$x=-\frac{4}{5}$$

Now solve the equation x=\frac{4±4}{-10} when ± is minus. Subtract 4 from 4.

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

Divide 0 by -10.

$$x=0$$

The equation is now solved.

$$x=-\frac{4}{5} x=0$$

Steps for Completing the Square

Multiply both sides of the equation by 5.

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

Subtract 5x from both sides.

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

Combine x and -5x to get -4x.

$$-4x=5x^{2}$$

Subtract 5x^{2} from both sides.

$$-4x-5x^{2}=0$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-5x^{2}-4x=0$$

Divide both sides by -5.

$$\frac{-5x^{2}-4x}{-5}=\frac{0}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\left(-\frac{4}{-5}\right)x=\frac{0}{-5}$$

Divide -4 by -5.

$$x^{2}+\frac{4}{5}x=\frac{0}{-5}$$

Divide 0 by -5.

$$x^{2}+\frac{4}{5}x=0$$

Divide \frac{4}{5}, the coefficient of the x term, by 2 to get \frac{2}{5}. Then add the square of \frac{2}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{4}{5}x+\left(\frac{2}{5}\right)^{2}=\left(\frac{2}{5}\right)^{2}$$

Square \frac{2}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{4}{5}x+\frac{4}{25}=\frac{4}{25}$$

Factor x^{2}+\frac{4}{5}x+\frac{4}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{2}{5}\right)^{2}=\frac{4}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{2}{5}\right)^{2}}=\sqrt{\frac{4}{25}}$$

Simplify.

$$x+\frac{2}{5}=\frac{2}{5} x+\frac{2}{5}=-\frac{2}{5}$$

Subtract \frac{2}{5} from both sides of the equation.

$$x=0 x=-\frac{4}{5}$$