Solve (5x)(8x-1)=90

$$(5x)(8x-1)=90$$

Solve for x

$$x = \frac{\sqrt{577} + 1}{16} \approx 1.563801519$$

$$x=\frac{1-\sqrt{577}}{16}\approx -1.438801519$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 5x by 8x-1.

$$40x^{2}-5x=90$$

Subtract 90 from both sides.

$$40x^{2}-5x-90=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 40 for a, -5 for b, and -90 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 40\left(-90\right)}}{2\times 40}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 40\left(-90\right)}}{2\times 40}$$

Multiply -4 times 40.

$$x=\frac{-\left(-5\right)±\sqrt{25-160\left(-90\right)}}{2\times 40}$$

Multiply -160 times -90.

$$x=\frac{-\left(-5\right)±\sqrt{25+14400}}{2\times 40}$$

Add 25 to 14400.

$$x=\frac{-\left(-5\right)±\sqrt{14425}}{2\times 40}$$

Take the square root of 14425.

$$x=\frac{-\left(-5\right)±5\sqrt{577}}{2\times 40}$$

The opposite of -5 is 5.

$$x=\frac{5±5\sqrt{577}}{2\times 40}$$

Multiply 2 times 40.

$$x=\frac{5±5\sqrt{577}}{80}$$

Now solve the equation x=\frac{5±5\sqrt{577}}{80} when ± is plus. Add 5 to 5\sqrt{577}.

$$x=\frac{5\sqrt{577}+5}{80}$$

Divide 5+5\sqrt{577} by 80.

$$x=\frac{\sqrt{577}+1}{16}$$

Now solve the equation x=\frac{5±5\sqrt{577}}{80} when ± is minus. Subtract 5\sqrt{577} from 5.

$$x=\frac{5-5\sqrt{577}}{80}$$

Divide 5-5\sqrt{577} by 80.

$$x=\frac{1-\sqrt{577}}{16}$$

The equation is now solved.

$$x=\frac{\sqrt{577}+1}{16} x=\frac{1-\sqrt{577}}{16}$$

Steps for Completing the Square

Use the distributive property to multiply 5x by 8x-1.

$$40x^{2}-5x=90$$

Divide both sides by 40.

$$\frac{40x^{2}-5x}{40}=\frac{90}{40}$$

Dividing by 40 undoes the multiplication by 40.

$$x^{2}+\left(-\frac{5}{40}\right)x=\frac{90}{40}$$

Reduce the fraction \frac{-5}{40} to lowest terms by extracting and canceling out 5.

$$x^{2}-\frac{1}{8}x=\frac{90}{40}$$

Reduce the fraction \frac{90}{40} to lowest terms by extracting and canceling out 10.

$$x^{2}-\frac{1}{8}x=\frac{9}{4}$$

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.

$$x^{2}-\frac{1}{8}x+\left(-\frac{1}{16}\right)^{2}=\frac{9}{4}+\left(-\frac{1}{16}\right)^{2}$$

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

$$x^{2}-\frac{1}{8}x+\frac{1}{256}=\frac{9}{4}+\frac{1}{256}$$

Add \frac{9}{4} to \frac{1}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{8}x+\frac{1}{256}=\frac{577}{256}$$

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}.

$$\left(x-\frac{1}{16}\right)^{2}=\frac{577}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{16}\right)^{2}}=\sqrt{\frac{577}{256}}$$

Simplify.

$$x-\frac{1}{16}=\frac{\sqrt{577}}{16} x-\frac{1}{16}=-\frac{\sqrt{577}}{16}$$

Add \frac{1}{16} to both sides of the equation.

$$x=\frac{\sqrt{577}+1}{16} x=\frac{1-\sqrt{577}}{16}$$