Solve -5=8/15x^2+1/15x-5

$$- 5 = \frac { 8 } { 15 } x ^ { 2 } + \frac { 1 } { 15 } x - 5$$

Solve for x

$$x=-\frac{1}{8}=-0.125$$

$$x=0$$

Steps Using Factoring

Swap sides so that all variable terms are on the left hand side.

$$\frac{8}{15}x^{2}+\frac{1}{15}x-5=-5$$

Add 5 to both sides.

$$\frac{8}{15}x^{2}+\frac{1}{15}x-5+5=0$$

Add -5 and 5 to get 0.

$$\frac{8}{15}x^{2}+\frac{1}{15}x=0$$

Factor out x.

$$x\left(\frac{8}{15}x+\frac{1}{15}\right)=0$$

To find equation solutions, solve x=0 and \frac{8x+1}{15}=0.

$$x=0 x=-\frac{1}{8}$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$\frac{8}{15}x^{2}+\frac{1}{15}x-5=-5$$

Add 5 to both sides.

$$\frac{8}{15}x^{2}+\frac{1}{15}x-5+5=0$$

Add -5 and 5 to get 0.

$$\frac{8}{15}x^{2}+\frac{1}{15}x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{8}{15} for a, \frac{1}{15} for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{1}{15}±\sqrt{\left(\frac{1}{15}\right)^{2}}}{2\times \frac{8}{15}}$$

Take the square root of \left(\frac{1}{15}\right)^{2}.

$$x=\frac{-\frac{1}{15}±\frac{1}{15}}{2\times \frac{8}{15}}$$

Multiply 2 times \frac{8}{15}.

$$x=\frac{-\frac{1}{15}±\frac{1}{15}}{\frac{16}{15}}$$

Now solve the equation x=\frac{-\frac{1}{15}±\frac{1}{15}}{\frac{16}{15}} when ± is plus. Add -\frac{1}{15} to \frac{1}{15} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{0}{\frac{16}{15}}$$

Divide 0 by \frac{16}{15} by multiplying 0 by the reciprocal of \frac{16}{15}.

$$x=0$$

Now solve the equation x=\frac{-\frac{1}{15}±\frac{1}{15}}{\frac{16}{15}} when ± is minus. Subtract \frac{1}{15} from -\frac{1}{15} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{\frac{2}{15}}{\frac{16}{15}}$$

Divide -\frac{2}{15} by \frac{16}{15} by multiplying -\frac{2}{15} by the reciprocal of \frac{16}{15}.

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

The equation is now solved.

$$x=0 x=-\frac{1}{8}$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$\frac{8}{15}x^{2}+\frac{1}{15}x-5=-5$$

Add 5 to both sides.

$$\frac{8}{15}x^{2}+\frac{1}{15}x=-5+5$$

Add -5 and 5 to get 0.

$$\frac{8}{15}x^{2}+\frac{1}{15}x=0$$

Divide both sides of the equation by \frac{8}{15}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{8}{15}x^{2}+\frac{1}{15}x}{\frac{8}{15}}=\frac{0}{\frac{8}{15}}$$

Dividing by \frac{8}{15} undoes the multiplication by \frac{8}{15}.

$$x^{2}+\frac{\frac{1}{15}}{\frac{8}{15}}x=\frac{0}{\frac{8}{15}}$$

Divide \frac{1}{15} by \frac{8}{15} by multiplying \frac{1}{15} by the reciprocal of \frac{8}{15}.

$$x^{2}+\frac{1}{8}x=\frac{0}{\frac{8}{15}}$$

Divide 0 by \frac{8}{15} by multiplying 0 by the reciprocal of \frac{8}{15}.

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

Divide \frac{1}{8}, the coefficient of the x term, by 2 to get \frac{1}{16}. Then add the square of \frac{1}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{8}x+\left(\frac{1}{16}\right)^{2}=\left(\frac{1}{16}\right)^{2}$$

Square \frac{1}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{8}x+\frac{1}{256}=\frac{1}{256}$$

Factor x^{2}+\frac{1}{8}x+\frac{1}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{16}\right)^{2}=\frac{1}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{16}\right)^{2}}=\sqrt{\frac{1}{256}}$$

Simplify.

$$x+\frac{1}{16}=\frac{1}{16} x+\frac{1}{16}=-\frac{1}{16}$$

Subtract \frac{1}{16} from both sides of the equation.

$$x=0 x=-\frac{1}{8}$$