Solve (500-2x)(10+x)=6000

$$(500-2x)(10+x)=6000$$

Solve for x

$$x=10\sqrt{139}+120\approx 237.898261226$$

$$x=120-10\sqrt{139}\approx 2.101738774$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 500-2x by 10+x and combine like terms.

$$5000+480x-2x^{2}=6000$$

Subtract 6000 from both sides.

$$5000+480x-2x^{2}-6000=0$$

Subtract 6000 from 5000 to get -1000.

$$-1000+480x-2x^{2}=0$$

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.

$$-2x^{2}+480x-1000=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 480 for b, and -1000 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-480±\sqrt{480^{2}-4\left(-2\right)\left(-1000\right)}}{2\left(-2\right)}$$

Square 480.

$$x=\frac{-480±\sqrt{230400-4\left(-2\right)\left(-1000\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-480±\sqrt{230400+8\left(-1000\right)}}{2\left(-2\right)}$$

Multiply 8 times -1000.

$$x=\frac{-480±\sqrt{230400-8000}}{2\left(-2\right)}$$

Add 230400 to -8000.

$$x=\frac{-480±\sqrt{222400}}{2\left(-2\right)}$$

Take the square root of 222400.

$$x=\frac{-480±40\sqrt{139}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-480±40\sqrt{139}}{-4}$$

Now solve the equation x=\frac{-480±40\sqrt{139}}{-4} when ± is plus. Add -480 to 40\sqrt{139}.

$$x=\frac{40\sqrt{139}-480}{-4}$$

Divide -480+40\sqrt{139} by -4.

$$x=120-10\sqrt{139}$$

Now solve the equation x=\frac{-480±40\sqrt{139}}{-4} when ± is minus. Subtract 40\sqrt{139} from -480.

$$x=\frac{-40\sqrt{139}-480}{-4}$$

Divide -480-40\sqrt{139} by -4.

$$x=10\sqrt{139}+120$$

The equation is now solved.

$$x=120-10\sqrt{139} x=10\sqrt{139}+120$$

Steps for Completing the Square

Use the distributive property to multiply 500-2x by 10+x and combine like terms.

$$5000+480x-2x^{2}=6000$$

Subtract 5000 from both sides.

$$480x-2x^{2}=6000-5000$$

Subtract 5000 from 6000 to get 1000.

$$480x-2x^{2}=1000$$

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.

$$-2x^{2}+480x=1000$$

Divide both sides by -2.

$$\frac{-2x^{2}+480x}{-2}=\frac{1000}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{480}{-2}x=\frac{1000}{-2}$$

Divide 480 by -2.

$$x^{2}-240x=\frac{1000}{-2}$$

Divide 1000 by -2.

$$x^{2}-240x=-500$$

Divide -240, the coefficient of the x term, by 2 to get -120. Then add the square of -120 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-240x+\left(-120\right)^{2}=-500+\left(-120\right)^{2}$$

Square -120.

$$x^{2}-240x+14400=-500+14400$$

Add -500 to 14400.

$$x^{2}-240x+14400=13900$$

Factor x^{2}-240x+14400. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-120\right)^{2}=13900$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-120\right)^{2}}=\sqrt{13900}$$

Simplify.

$$x-120=10\sqrt{139} x-120=-10\sqrt{139}$$

Add 120 to both sides of the equation.

$$x=10\sqrt{139}+120 x=120-10\sqrt{139}$$