Solve (2x-3)(x-4)=18

$$(2x-3)(x-4)=18$$

Solve for x

$$x=-\frac{1}{2}=-0.5$$

$$x=6$$

Steps Using the Quadratic Formula

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

$$2x^{2}-11x+12=18$$

Subtract 18 from both sides.

$$2x^{2}-11x+12-18=0$$

Subtract 18 from 12 to get -6.

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

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

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

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 2\left(-6\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-11\right)±\sqrt{121-8\left(-6\right)}}{2\times 2}$$

Multiply -8 times -6.

$$x=\frac{-\left(-11\right)±\sqrt{121+48}}{2\times 2}$$

Add 121 to 48.

$$x=\frac{-\left(-11\right)±\sqrt{169}}{2\times 2}$$

Take the square root of 169.

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

The opposite of -11 is 11.

$$x=\frac{11±13}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{11±13}{4}$$

Now solve the equation x=\frac{11±13}{4} when ± is plus. Add 11 to 13.

$$x=\frac{24}{4}$$

Divide 24 by 4.

$$x=6$$

Now solve the equation x=\frac{11±13}{4} when ± is minus. Subtract 13 from 11.

$$x=-\frac{2}{4}$$

Reduce the fraction \frac{-2}{4} to lowest terms by extracting and canceling out 2.

$$x=-\frac{1}{2}$$

The equation is now solved.

$$x=6 x=-\frac{1}{2}$$

Steps for Completing the Square

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

$$2x^{2}-11x+12=18$$

Subtract 12 from both sides.

$$2x^{2}-11x=18-12$$

Subtract 12 from 18 to get 6.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide 6 by 2.

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

Divide -\frac{11}{2}, the coefficient of the x term, by 2 to get -\frac{11}{4}. Then add the square of -\frac{11}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{2}x+\left(-\frac{11}{4}\right)^{2}=3+\left(-\frac{11}{4}\right)^{2}$$

Square -\frac{11}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=3+\frac{121}{16}$$

Add 3 to \frac{121}{16}.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=\frac{169}{16}$$

Factor x^{2}-\frac{11}{2}x+\frac{121}{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}.

$$\left(x-\frac{11}{4}\right)^{2}=\frac{169}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{4}\right)^{2}}=\sqrt{\frac{169}{16}}$$

Simplify.

$$x-\frac{11}{4}=\frac{13}{4} x-\frac{11}{4}=-\frac{13}{4}$$

Add \frac{11}{4} to both sides of the equation.

$$x=6 x=-\frac{1}{2}$$