Solve (20-2x)(100+20x)=2240

$$(20-2x)(100+20x)=2240$$

Solve for x

$$x=2$$

$$x=3$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 20-2x by 100+20x and combine like terms.

$$2000+200x-40x^{2}=2240$$

Subtract 2240 from both sides.

$$2000+200x-40x^{2}-2240=0$$

Subtract 2240 from 2000 to get -240.

$$-240+200x-40x^{2}=0$$

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.

$$-40x^{2}+200x-240=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -40 for a, 200 for b, and -240 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-200±\sqrt{200^{2}-4\left(-40\right)\left(-240\right)}}{2\left(-40\right)}$$

Square 200.

$$x=\frac{-200±\sqrt{40000-4\left(-40\right)\left(-240\right)}}{2\left(-40\right)}$$

Multiply -4 times -40.

$$x=\frac{-200±\sqrt{40000+160\left(-240\right)}}{2\left(-40\right)}$$

Multiply 160 times -240.

$$x=\frac{-200±\sqrt{40000-38400}}{2\left(-40\right)}$$

Add 40000 to -38400.

$$x=\frac{-200±\sqrt{1600}}{2\left(-40\right)}$$

Take the square root of 1600.

$$x=\frac{-200±40}{2\left(-40\right)}$$

Multiply 2 times -40.

$$x=\frac{-200±40}{-80}$$

Now solve the equation x=\frac{-200±40}{-80} when ± is plus. Add -200 to 40.

$$x=-\frac{160}{-80}$$

Divide -160 by -80.

$$x=2$$

Now solve the equation x=\frac{-200±40}{-80} when ± is minus. Subtract 40 from -200.

$$x=-\frac{240}{-80}$$

Divide -240 by -80.

$$x=3$$

The equation is now solved.

$$x=2 x=3$$

Steps for Completing the Square

Use the distributive property to multiply 20-2x by 100+20x and combine like terms.

$$2000+200x-40x^{2}=2240$$

Subtract 2000 from both sides.

$$200x-40x^{2}=2240-2000$$

Subtract 2000 from 2240 to get 240.

$$200x-40x^{2}=240$$

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.

$$-40x^{2}+200x=240$$

Divide both sides by -40.

$$\frac{-40x^{2}+200x}{-40}=\frac{240}{-40}$$

Dividing by -40 undoes the multiplication by -40.

$$x^{2}+\frac{200}{-40}x=\frac{240}{-40}$$

Divide 200 by -40.

$$x^{2}-5x=\frac{240}{-40}$$

Divide 240 by -40.

$$x^{2}-5x=-6$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-6+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=-6+\frac{25}{4}$$

Add -6 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{1}{4}$$

Factor x^{2}-5x+\frac{25}{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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{1}{2} x-\frac{5}{2}=-\frac{1}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=3 x=2$$