Solve (3x-4)(2x-5)=31

$$( 3 x - 4 ) ( 2 x - 5 ) = 31$$

Solve for x

$$x = \frac{\sqrt{793} + 23}{12} \approx 4.26335464$$

$$x=\frac{23-\sqrt{793}}{12}\approx -0.430021307$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x-4 by 2x-5 and combine like terms.

$$6x^{2}-23x+20=31$$

Subtract 31 from both sides.

$$6x^{2}-23x+20-31=0$$

Subtract 31 from 20 to get -11.

$$6x^{2}-23x-11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -23 for b, and -11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-23\right)±\sqrt{\left(-23\right)^{2}-4\times 6\left(-11\right)}}{2\times 6}$$

Square -23.

$$x=\frac{-\left(-23\right)±\sqrt{529-4\times 6\left(-11\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-23\right)±\sqrt{529-24\left(-11\right)}}{2\times 6}$$

Multiply -24 times -11.

$$x=\frac{-\left(-23\right)±\sqrt{529+264}}{2\times 6}$$

Add 529 to 264.

$$x=\frac{-\left(-23\right)±\sqrt{793}}{2\times 6}$$

The opposite of -23 is 23.

$$x=\frac{23±\sqrt{793}}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{23±\sqrt{793}}{12}$$

Now solve the equation x=\frac{23±\sqrt{793}}{12} when ± is plus. Add 23 to \sqrt{793}.

$$x=\frac{\sqrt{793}+23}{12}$$

Now solve the equation x=\frac{23±\sqrt{793}}{12} when ± is minus. Subtract \sqrt{793} from 23.

$$x=\frac{23-\sqrt{793}}{12}$$

The equation is now solved.

$$x=\frac{\sqrt{793}+23}{12} x=\frac{23-\sqrt{793}}{12}$$

Steps for Completing the Square

Use the distributive property to multiply 3x-4 by 2x-5 and combine like terms.

$$6x^{2}-23x+20=31$$

Subtract 20 from both sides.

$$6x^{2}-23x=31-20$$

Subtract 20 from 31 to get 11.

$$6x^{2}-23x=11$$

Divide both sides by 6.

$$\frac{6x^{2}-23x}{6}=\frac{11}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}-\frac{23}{6}x=\frac{11}{6}$$

Divide -\frac{23}{6}, the coefficient of the x term, by 2 to get -\frac{23}{12}. Then add the square of -\frac{23}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{23}{6}x+\left(-\frac{23}{12}\right)^{2}=\frac{11}{6}+\left(-\frac{23}{12}\right)^{2}$$

Square -\frac{23}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{23}{6}x+\frac{529}{144}=\frac{11}{6}+\frac{529}{144}$$

Add \frac{11}{6} to \frac{529}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{23}{6}x+\frac{529}{144}=\frac{793}{144}$$

Factor x^{2}-\frac{23}{6}x+\frac{529}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{23}{12}\right)^{2}=\frac{793}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{23}{12}\right)^{2}}=\sqrt{\frac{793}{144}}$$

Simplify.

$$x-\frac{23}{12}=\frac{\sqrt{793}}{12} x-\frac{23}{12}=-\frac{\sqrt{793}}{12}$$

Add \frac{23}{12} to both sides of the equation.

$$x=\frac{\sqrt{793}+23}{12} x=\frac{23-\sqrt{793}}{12}$$