Solve (20-x)(50+2x)=700

$$(20-x)(50+2x)=700$$

Solve for x

$$x=-15$$

$$x=10$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 20-x by 50+2x and combine like terms.

$$1000-10x-2x^{2}=700$$

Subtract 700 from both sides.

$$1000-10x-2x^{2}-700=0$$

Subtract 700 from 1000 to get 300.

$$300-10x-2x^{2}=0$$

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.

$$-2x^{2}-10x+300=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -10 for b, and 300 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-2\right)\times 300}}{2\left(-2\right)}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-2\right)\times 300}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-10\right)±\sqrt{100+8\times 300}}{2\left(-2\right)}$$

Multiply 8 times 300.

$$x=\frac{-\left(-10\right)±\sqrt{100+2400}}{2\left(-2\right)}$$

Add 100 to 2400.

$$x=\frac{-\left(-10\right)±\sqrt{2500}}{2\left(-2\right)}$$

Take the square root of 2500.

$$x=\frac{-\left(-10\right)±50}{2\left(-2\right)}$$

The opposite of -10 is 10.

$$x=\frac{10±50}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{10±50}{-4}$$

Now solve the equation x=\frac{10±50}{-4} when ± is plus. Add 10 to 50.

$$x=\frac{60}{-4}$$

Divide 60 by -4.

$$x=-15$$

Now solve the equation x=\frac{10±50}{-4} when ± is minus. Subtract 50 from 10.

$$x=-\frac{40}{-4}$$

Divide -40 by -4.

$$x=10$$

The equation is now solved.

$$x=-15 x=10$$

Steps for Completing the Square

Use the distributive property to multiply 20-x by 50+2x and combine like terms.

$$1000-10x-2x^{2}=700$$

Subtract 1000 from both sides.

$$-10x-2x^{2}=700-1000$$

Subtract 1000 from 700 to get -300.

$$-10x-2x^{2}=-300$$

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.

$$-2x^{2}-10x=-300$$

Divide both sides by -2.

$$\frac{-2x^{2}-10x}{-2}=-\frac{300}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\left(-\frac{10}{-2}\right)x=-\frac{300}{-2}$$

Divide -10 by -2.

$$x^{2}+5x=-\frac{300}{-2}$$

Divide -300 by -2.

$$x^{2}+5x=150$$

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.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=150+\left(\frac{5}{2}\right)^{2}$$

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

$$x^{2}+5x+\frac{25}{4}=150+\frac{25}{4}$$

Add 150 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{625}{4}$$

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}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{625}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{625}{4}}$$

Simplify.

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

Subtract \frac{5}{2} from both sides of the equation.

$$x=10 x=-15$$