Solve (3x-11)(x-2)=2

$$( 3 x - 11 ) ( x - 2 ) = 2$$

Solve for x

$$x=4$$

$$x = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x-11 by x-2 and combine like terms.

$$3x^{2}-17x+22=2$$

Subtract 2 from both sides.

$$3x^{2}-17x+22-2=0$$

Subtract 2 from 22 to get 20.

$$3x^{2}-17x+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -17 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 3\times 20}}{2\times 3}$$

Square -17.

$$x=\frac{-\left(-17\right)±\sqrt{289-4\times 3\times 20}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-17\right)±\sqrt{289-12\times 20}}{2\times 3}$$

Multiply -12 times 20.

$$x=\frac{-\left(-17\right)±\sqrt{289-240}}{2\times 3}$$

Add 289 to -240.

$$x=\frac{-\left(-17\right)±\sqrt{49}}{2\times 3}$$

Take the square root of 49.

$$x=\frac{-\left(-17\right)±7}{2\times 3}$$

The opposite of -17 is 17.

$$x=\frac{17±7}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{17±7}{6}$$

Now solve the equation x=\frac{17±7}{6} when ± is plus. Add 17 to 7.

$$x=\frac{24}{6}$$

Divide 24 by 6.

$$x=4$$

Now solve the equation x=\frac{17±7}{6} when ± is minus. Subtract 7 from 17.

$$x=\frac{10}{6}$$

Reduce the fraction \frac{10}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{3}$$

The equation is now solved.

$$x=4 x=\frac{5}{3}$$

Steps for Completing the Square

Use the distributive property to multiply 3x-11 by x-2 and combine like terms.

$$3x^{2}-17x+22=2$$

Subtract 22 from both sides.

$$3x^{2}-17x=2-22$$

Subtract 22 from 2 to get -20.

$$3x^{2}-17x=-20$$

Divide both sides by 3.

$$\frac{3x^{2}-17x}{3}=-\frac{20}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{17}{3}x=-\frac{20}{3}$$

Divide -\frac{17}{3}, the coefficient of the x term, by 2 to get -\frac{17}{6}. Then add the square of -\frac{17}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{17}{3}x+\left(-\frac{17}{6}\right)^{2}=-\frac{20}{3}+\left(-\frac{17}{6}\right)^{2}$$

Square -\frac{17}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{17}{3}x+\frac{289}{36}=-\frac{20}{3}+\frac{289}{36}$$

Add -\frac{20}{3} to \frac{289}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{17}{3}x+\frac{289}{36}=\frac{49}{36}$$

Factor x^{2}-\frac{17}{3}x+\frac{289}{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}.

$$\left(x-\frac{17}{6}\right)^{2}=\frac{49}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{17}{6}\right)^{2}}=\sqrt{\frac{49}{36}}$$

Simplify.

$$x-\frac{17}{6}=\frac{7}{6} x-\frac{17}{6}=-\frac{7}{6}$$

Add \frac{17}{6} to both sides of the equation.

$$x=4 x=\frac{5}{3}$$