Solve (x-2)(x-1)=40

$$(x-2)(x-1)=40$$

Solve for x

$$x = \frac{\sqrt{161} + 3}{2} \approx 7.84428877$$

$$x=\frac{3-\sqrt{161}}{2}\approx -4.84428877$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x-2 by x-1 and combine like terms.

$$x^{2}-3x+2=40$$

Subtract 40 from both sides.

$$x^{2}-3x+2-40=0$$

Subtract 40 from 2 to get -38.

$$x^{2}-3x-38=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and -38 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-38\right)}}{2}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-38\right)}}{2}$$

Multiply -4 times -38.

$$x=\frac{-\left(-3\right)±\sqrt{9+152}}{2}$$

Add 9 to 152.

$$x=\frac{-\left(-3\right)±\sqrt{161}}{2}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{161}}{2}$$

Now solve the equation x=\frac{3±\sqrt{161}}{2} when ± is plus. Add 3 to \sqrt{161}.

$$x=\frac{\sqrt{161}+3}{2}$$

Now solve the equation x=\frac{3±\sqrt{161}}{2} when ± is minus. Subtract \sqrt{161} from 3.

$$x=\frac{3-\sqrt{161}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{161}+3}{2} x=\frac{3-\sqrt{161}}{2}$$

Steps for Completing the Square

Use the distributive property to multiply x-2 by x-1 and combine like terms.

$$x^{2}-3x+2=40$$

Subtract 2 from both sides.

$$x^{2}-3x=40-2$$

Subtract 2 from 40 to get 38.

$$x^{2}-3x=38$$

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

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=38+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=38+\frac{9}{4}$$

Add 38 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{161}{4}$$

Factor x^{2}-3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{161}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{161}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{161}}{2} x-\frac{3}{2}=-\frac{\sqrt{161}}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{\sqrt{161}+3}{2} x=\frac{3-\sqrt{161}}{2}$$