Solve (x-2)(3x-5)=1

$$(x-2)(3x-5)=1$$

Solve for x

$$x = \frac{\sqrt{13} + 11}{6} \approx 2.434258546$$

$$x = \frac{11 - \sqrt{13}}{6} \approx 1.232408121$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x-2 by 3x-5 and combine like terms.

$$3x^{2}-11x+10=1$$

Subtract 1 from both sides.

$$3x^{2}-11x+10-1=0$$

Subtract 1 from 10 to get 9.

$$3x^{2}-11x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -11 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 3\times 9}}{2\times 3}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 3\times 9}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-11\right)±\sqrt{121-12\times 9}}{2\times 3}$$

Multiply -12 times 9.

$$x=\frac{-\left(-11\right)±\sqrt{121-108}}{2\times 3}$$

Add 121 to -108.

$$x=\frac{-\left(-11\right)±\sqrt{13}}{2\times 3}$$

The opposite of -11 is 11.

$$x=\frac{11±\sqrt{13}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{11±\sqrt{13}}{6}$$

Now solve the equation x=\frac{11±\sqrt{13}}{6} when ± is plus. Add 11 to \sqrt{13}.

$$x=\frac{\sqrt{13}+11}{6}$$

Now solve the equation x=\frac{11±\sqrt{13}}{6} when ± is minus. Subtract \sqrt{13} from 11.

$$x=\frac{11-\sqrt{13}}{6}$$

The equation is now solved.

$$x=\frac{\sqrt{13}+11}{6} x=\frac{11-\sqrt{13}}{6}$$

Steps for Completing the Square

Use the distributive property to multiply x-2 by 3x-5 and combine like terms.

$$3x^{2}-11x+10=1$$

Subtract 10 from both sides.

$$3x^{2}-11x=1-10$$

Subtract 10 from 1 to get -9.

$$3x^{2}-11x=-9$$

Divide both sides by 3.

$$\frac{3x^{2}-11x}{3}=-\frac{9}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{11}{3}x=-\frac{9}{3}$$

Divide -9 by 3.

$$x^{2}-\frac{11}{3}x=-3$$

Divide -\frac{11}{3}, the coefficient of the x term, by 2 to get -\frac{11}{6}. Then add the square of -\frac{11}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{3}x+\left(-\frac{11}{6}\right)^{2}=-3+\left(-\frac{11}{6}\right)^{2}$$

Square -\frac{11}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{3}x+\frac{121}{36}=-3+\frac{121}{36}$$

Add -3 to \frac{121}{36}.

$$x^{2}-\frac{11}{3}x+\frac{121}{36}=\frac{13}{36}$$

Factor x^{2}-\frac{11}{3}x+\frac{121}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{6}\right)^{2}=\frac{13}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{6}\right)^{2}}=\sqrt{\frac{13}{36}}$$

Simplify.

$$x-\frac{11}{6}=\frac{\sqrt{13}}{6} x-\frac{11}{6}=-\frac{\sqrt{13}}{6}$$

Add \frac{11}{6} to both sides of the equation.

$$x=\frac{\sqrt{13}+11}{6} x=\frac{11-\sqrt{13}}{6}$$