Solve 5x^2=x+8

$$5 { x }^{ 2 } =x+8$$

Solve for x

$$x = \frac{\sqrt{161} + 1}{10} \approx 1.368857754$$

$$x=\frac{1-\sqrt{161}}{10}\approx -1.168857754$$

Steps Using the Quadratic Formula

Subtract x from both sides.

$$5x^{2}-x=8$$

Subtract 8 from both sides.

$$5x^{2}-x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -1 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 5\left(-8\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-1\right)±\sqrt{1-20\left(-8\right)}}{2\times 5}$$

Multiply -20 times -8.

$$x=\frac{-\left(-1\right)±\sqrt{1+160}}{2\times 5}$$

Add 1 to 160.

$$x=\frac{-\left(-1\right)±\sqrt{161}}{2\times 5}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{161}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{1±\sqrt{161}}{10}$$

Now solve the equation x=\frac{1±\sqrt{161}}{10} when ± is plus. Add 1 to \sqrt{161}.

$$x=\frac{\sqrt{161}+1}{10}$$

Now solve the equation x=\frac{1±\sqrt{161}}{10} when ± is minus. Subtract \sqrt{161} from 1.

$$x=\frac{1-\sqrt{161}}{10}$$

The equation is now solved.

$$x=\frac{\sqrt{161}+1}{10} x=\frac{1-\sqrt{161}}{10}$$

Steps for Completing the Square

Subtract x from both sides.

$$5x^{2}-x=8$$

Divide both sides by 5.

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

Dividing by 5 undoes the multiplication by 5.

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

Divide -\frac{1}{5}, the coefficient of the x term, by 2 to get -\frac{1}{10}. Then add the square of -\frac{1}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{5}x+\left(-\frac{1}{10}\right)^{2}=\frac{8}{5}+\left(-\frac{1}{10}\right)^{2}$$

Square -\frac{1}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{8}{5}+\frac{1}{100}$$

Add \frac{8}{5} to \frac{1}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{161}{100}$$

Factor x^{2}-\frac{1}{5}x+\frac{1}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{10}\right)^{2}=\frac{161}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{10}\right)^{2}}=\sqrt{\frac{161}{100}}$$

Simplify.

$$x-\frac{1}{10}=\frac{\sqrt{161}}{10} x-\frac{1}{10}=-\frac{\sqrt{161}}{10}$$

Add \frac{1}{10} to both sides of the equation.

$$x=\frac{\sqrt{161}+1}{10} x=\frac{1-\sqrt{161}}{10}$$