Solve (40-m)(20+2m)=120

$$( 40 - m ) ( 20 + 2 m ) = 120$$

Solve for m

$$m=\sqrt{565}+15\approx 38.769728648$$

$$m=15-\sqrt{565}\approx -8.769728648$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 40-m by 20+2m and combine like terms.

$$800+60m-2m^{2}=120$$

Subtract 120 from both sides.

$$800+60m-2m^{2}-120=0$$

Subtract 120 from 800 to get 680.

$$680+60m-2m^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-2m^{2}+60m+680=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 60 for b, and 680 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$m=\frac{-60±\sqrt{60^{2}-4\left(-2\right)\times 680}}{2\left(-2\right)}$$

Square 60.

$$m=\frac{-60±\sqrt{3600-4\left(-2\right)\times 680}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$m=\frac{-60±\sqrt{3600+8\times 680}}{2\left(-2\right)}$$

Multiply 8 times 680.

$$m=\frac{-60±\sqrt{3600+5440}}{2\left(-2\right)}$$

Add 3600 to 5440.

$$m=\frac{-60±\sqrt{9040}}{2\left(-2\right)}$$

Take the square root of 9040.

$$m=\frac{-60±4\sqrt{565}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$m=\frac{-60±4\sqrt{565}}{-4}$$

Now solve the equation m=\frac{-60±4\sqrt{565}}{-4} when ± is plus. Add -60 to 4\sqrt{565}.

$$m=\frac{4\sqrt{565}-60}{-4}$$

Divide -60+4\sqrt{565} by -4.

$$m=15-\sqrt{565}$$

Now solve the equation m=\frac{-60±4\sqrt{565}}{-4} when ± is minus. Subtract 4\sqrt{565} from -60.

$$m=\frac{-4\sqrt{565}-60}{-4}$$

Divide -60-4\sqrt{565} by -4.

$$m=\sqrt{565}+15$$

The equation is now solved.

$$m=15-\sqrt{565} m=\sqrt{565}+15$$

Steps for Completing the Square

Use the distributive property to multiply 40-m by 20+2m and combine like terms.

$$800+60m-2m^{2}=120$$

Subtract 800 from both sides.

$$60m-2m^{2}=120-800$$

Subtract 800 from 120 to get -680.

$$60m-2m^{2}=-680$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-2m^{2}+60m=-680$$

Divide both sides by -2.

$$\frac{-2m^{2}+60m}{-2}=-\frac{680}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$m^{2}+\frac{60}{-2}m=-\frac{680}{-2}$$

Divide 60 by -2.

$$m^{2}-30m=-\frac{680}{-2}$$

Divide -680 by -2.

$$m^{2}-30m=340$$

Divide -30, the coefficient of the x term, by 2 to get -15. Then add the square of -15 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$m^{2}-30m+\left(-15\right)^{2}=340+\left(-15\right)^{2}$$

Square -15.

$$m^{2}-30m+225=340+225$$

Add 340 to 225.

$$m^{2}-30m+225=565$$

Factor m^{2}-30m+225. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(m-15\right)^{2}=565$$

Take the square root of both sides of the equation.

$$\sqrt{\left(m-15\right)^{2}}=\sqrt{565}$$

Simplify.

$$m-15=\sqrt{565} m-15=-\sqrt{565}$$

Add 15 to both sides of the equation.

$$m=\sqrt{565}+15 m=15-\sqrt{565}$$