Solve left(x-5right)left(2x-7right)=3

$$\left( x-5 \right) \left( 2x-7 \right) = 3$$

Solve for x

$$x = \frac{\sqrt{33} + 17}{4} \approx 5.686140662$$

$$x = \frac{17 - \sqrt{33}}{4} \approx 2.813859338$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x-5 by 2x-7 and combine like terms.

$$2x^{2}-17x+35=3$$

Subtract 3 from both sides.

$$2x^{2}-17x+35-3=0$$

Subtract 3 from 35 to get 32.

$$2x^{2}-17x+32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -17 for b, and 32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 2\times 32}}{2\times 2}$$

Square -17.

$$x=\frac{-\left(-17\right)±\sqrt{289-4\times 2\times 32}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-17\right)±\sqrt{289-8\times 32}}{2\times 2}$$

Multiply -8 times 32.

$$x=\frac{-\left(-17\right)±\sqrt{289-256}}{2\times 2}$$

Add 289 to -256.

$$x=\frac{-\left(-17\right)±\sqrt{33}}{2\times 2}$$

The opposite of -17 is 17.

$$x=\frac{17±\sqrt{33}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{17±\sqrt{33}}{4}$$

Now solve the equation x=\frac{17±\sqrt{33}}{4} when ± is plus. Add 17 to \sqrt{33}.

$$x=\frac{\sqrt{33}+17}{4}$$

Now solve the equation x=\frac{17±\sqrt{33}}{4} when ± is minus. Subtract \sqrt{33} from 17.

$$x=\frac{17-\sqrt{33}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{33}+17}{4} x=\frac{17-\sqrt{33}}{4}$$

Steps for Completing the Square

Use the distributive property to multiply x-5 by 2x-7 and combine like terms.

$$2x^{2}-17x+35=3$$

Subtract 35 from both sides.

$$2x^{2}-17x=3-35$$

Subtract 35 from 3 to get -32.

$$2x^{2}-17x=-32$$

Divide both sides by 2.

$$\frac{2x^{2}-17x}{2}=-\frac{32}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{17}{2}x=-\frac{32}{2}$$

Divide -32 by 2.

$$x^{2}-\frac{17}{2}x=-16$$

Divide -\frac{17}{2}, the coefficient of the x term, by 2 to get -\frac{17}{4}. Then add the square of -\frac{17}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{17}{2}x+\left(-\frac{17}{4}\right)^{2}=-16+\left(-\frac{17}{4}\right)^{2}$$

Square -\frac{17}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{17}{2}x+\frac{289}{16}=-16+\frac{289}{16}$$

Add -16 to \frac{289}{16}.

$$x^{2}-\frac{17}{2}x+\frac{289}{16}=\frac{33}{16}$$

Factor x^{2}-\frac{17}{2}x+\frac{289}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{17}{4}\right)^{2}=\frac{33}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{17}{4}\right)^{2}}=\sqrt{\frac{33}{16}}$$

Simplify.

$$x-\frac{17}{4}=\frac{\sqrt{33}}{4} x-\frac{17}{4}=-\frac{\sqrt{33}}{4}$$

Add \frac{17}{4} to both sides of the equation.

$$x=\frac{\sqrt{33}+17}{4} x=\frac{17-\sqrt{33}}{4}$$