Solve 25x^2=5x+8

$$25 { x }^{ 2 } =5x+8$$

Solve for x

$$x=\frac{\sqrt{33}+1}{10}\approx 0.674456265$$

$$x=\frac{1-\sqrt{33}}{10}\approx -0.474456265$$

Steps Using the Quadratic Formula

Subtract 5x from both sides.

$$25x^{2}-5x=8$$

Subtract 8 from both sides.

$$25x^{2}-5x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, -5 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 25\left(-8\right)}}{2\times 25}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 25\left(-8\right)}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{-\left(-5\right)±\sqrt{25-100\left(-8\right)}}{2\times 25}$$

Multiply -100 times -8.

$$x=\frac{-\left(-5\right)±\sqrt{25+800}}{2\times 25}$$

Add 25 to 800.

$$x=\frac{-\left(-5\right)±\sqrt{825}}{2\times 25}$$

Take the square root of 825.

$$x=\frac{-\left(-5\right)±5\sqrt{33}}{2\times 25}$$

The opposite of -5 is 5.

$$x=\frac{5±5\sqrt{33}}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{5±5\sqrt{33}}{50}$$

Now solve the equation x=\frac{5±5\sqrt{33}}{50} when ± is plus. Add 5 to 5\sqrt{33}.

$$x=\frac{5\sqrt{33}+5}{50}$$

Divide 5+5\sqrt{33} by 50.

$$x=\frac{\sqrt{33}+1}{10}$$

Now solve the equation x=\frac{5±5\sqrt{33}}{50} when ± is minus. Subtract 5\sqrt{33} from 5.

$$x=\frac{5-5\sqrt{33}}{50}$$

Divide 5-5\sqrt{33} by 50.

$$x=\frac{1-\sqrt{33}}{10}$$

The equation is now solved.

$$x=\frac{\sqrt{33}+1}{10} x=\frac{1-\sqrt{33}}{10}$$

Steps for Completing the Square

Subtract 5x from both sides.

$$25x^{2}-5x=8$$

Divide both sides by 25.

$$\frac{25x^{2}-5x}{25}=\frac{8}{25}$$

Dividing by 25 undoes the multiplication by 25.

$$x^{2}+\left(-\frac{5}{25}\right)x=\frac{8}{25}$$

Reduce the fraction \frac{-5}{25} to lowest terms by extracting and canceling out 5.

$$x^{2}-\frac{1}{5}x=\frac{8}{25}$$

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.

$$x^{2}-\frac{1}{5}x+\left(-\frac{1}{10}\right)^{2}=\frac{8}{25}+\left(-\frac{1}{10}\right)^{2}$$

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

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{8}{25}+\frac{1}{100}$$

Add \frac{8}{25} to \frac{1}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{33}{100}$$

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}.

$$\left(x-\frac{1}{10}\right)^{2}=\frac{33}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{10}\right)^{2}}=\sqrt{\frac{33}{100}}$$

Simplify.

$$x-\frac{1}{10}=\frac{\sqrt{33}}{10} x-\frac{1}{10}=-\frac{\sqrt{33}}{10}$$

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

$$x=\frac{\sqrt{33}+1}{10} x=\frac{1-\sqrt{33}}{10}$$