Solve (2x-1)*(4x-7)=1

$$(2x-1) \times (4x-7)=1$$

Solve for x

$$x = \frac{\sqrt{33} + 9}{8} \approx 1.843070331$$

$$x=\frac{9-\sqrt{33}}{8}\approx 0.406929669$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x-1 by 4x-7 and combine like terms.

$$8x^{2}-18x+7=1$$

Subtract 1 from both sides.

$$8x^{2}-18x+7-1=0$$

Subtract 1 from 7 to get 6.

$$8x^{2}-18x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, -18 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 8\times 6}}{2\times 8}$$

Square -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\times 8\times 6}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-\left(-18\right)±\sqrt{324-32\times 6}}{2\times 8}$$

Multiply -32 times 6.

$$x=\frac{-\left(-18\right)±\sqrt{324-192}}{2\times 8}$$

Add 324 to -192.

$$x=\frac{-\left(-18\right)±\sqrt{132}}{2\times 8}$$

Take the square root of 132.

$$x=\frac{-\left(-18\right)±2\sqrt{33}}{2\times 8}$$

The opposite of -18 is 18.

$$x=\frac{18±2\sqrt{33}}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{18±2\sqrt{33}}{16}$$

Now solve the equation x=\frac{18±2\sqrt{33}}{16} when ± is plus. Add 18 to 2\sqrt{33}.

$$x=\frac{2\sqrt{33}+18}{16}$$

Divide 18+2\sqrt{33} by 16.

$$x=\frac{\sqrt{33}+9}{8}$$

Now solve the equation x=\frac{18±2\sqrt{33}}{16} when ± is minus. Subtract 2\sqrt{33} from 18.

$$x=\frac{18-2\sqrt{33}}{16}$$

Divide 18-2\sqrt{33} by 16.

$$x=\frac{9-\sqrt{33}}{8}$$

The equation is now solved.

$$x=\frac{\sqrt{33}+9}{8} x=\frac{9-\sqrt{33}}{8}$$

Steps for Completing the Square

Use the distributive property to multiply 2x-1 by 4x-7 and combine like terms.

$$8x^{2}-18x+7=1$$

Subtract 7 from both sides.

$$8x^{2}-18x=1-7$$

Subtract 7 from 1 to get -6.

$$8x^{2}-18x=-6$$

Divide both sides by 8.

$$\frac{8x^{2}-18x}{8}=-\frac{6}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\left(-\frac{18}{8}\right)x=-\frac{6}{8}$$

Reduce the fraction \frac{-18}{8} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{9}{4}x=-\frac{6}{8}$$

Reduce the fraction \frac{-6}{8} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{9}{4}x=-\frac{3}{4}$$

Divide -\frac{9}{4}, the coefficient of the x term, by 2 to get -\frac{9}{8}. Then add the square of -\frac{9}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{9}{4}x+\left(-\frac{9}{8}\right)^{2}=-\frac{3}{4}+\left(-\frac{9}{8}\right)^{2}$$

Square -\frac{9}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{9}{4}x+\frac{81}{64}=-\frac{3}{4}+\frac{81}{64}$$

Add -\frac{3}{4} to \frac{81}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{9}{4}x+\frac{81}{64}=\frac{33}{64}$$

Factor x^{2}-\frac{9}{4}x+\frac{81}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{9}{8}\right)^{2}=\frac{33}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{8}\right)^{2}}=\sqrt{\frac{33}{64}}$$

Simplify.

$$x-\frac{9}{8}=\frac{\sqrt{33}}{8} x-\frac{9}{8}=-\frac{\sqrt{33}}{8}$$

Add \frac{9}{8} to both sides of the equation.

$$x=\frac{\sqrt{33}+9}{8} x=\frac{9-\sqrt{33}}{8}$$