Solve 500=60x-x^2

$$500=60x- { x }^{ 2 }$$

Solve for x

$$x=10$$

$$x=50$$

Steps Using Factoring By Grouping

Swap sides so that all variable terms are on the left hand side.

$$60x-x^{2}=500$$

Subtract 500 from both sides.

$$60x-x^{2}-500=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}+60x-500=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-500. To find a and b, set up a system to be solved.

$$a+b=60 ab=-\left(-500\right)=500$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 500.

$$1,500 2,250 4,125 5,100 10,50 20,25$$

Calculate the sum for each pair.

$$1+500=501 2+250=252 4+125=129 5+100=105 10+50=60 20+25=45$$

The solution is the pair that gives sum 60.

$$a=50 b=10$$

Rewrite -x^{2}+60x-500 as \left(-x^{2}+50x\right)+\left(10x-500\right).

$$\left(-x^{2}+50x\right)+\left(10x-500\right)$$

Factor out -x in the first and 10 in the second group.

$$-x\left(x-50\right)+10\left(x-50\right)$$

Factor out common term x-50 by using distributive property.

$$\left(x-50\right)\left(-x+10\right)$$

To find equation solutions, solve x-50=0 and -x+10=0.

$$x=50 x=10$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$60x-x^{2}=500$$

Subtract 500 from both sides.

$$60x-x^{2}-500=0$$

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.

$$-x^{2}+60x-500=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 60 for b, and -500 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-60±\sqrt{60^{2}-4\left(-1\right)\left(-500\right)}}{2\left(-1\right)}$$

Square 60.

$$x=\frac{-60±\sqrt{3600-4\left(-1\right)\left(-500\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-60±\sqrt{3600+4\left(-500\right)}}{2\left(-1\right)}$$

Multiply 4 times -500.

$$x=\frac{-60±\sqrt{3600-2000}}{2\left(-1\right)}$$

Add 3600 to -2000.

$$x=\frac{-60±\sqrt{1600}}{2\left(-1\right)}$$

Take the square root of 1600.

$$x=\frac{-60±40}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-60±40}{-2}$$

Now solve the equation x=\frac{-60±40}{-2} when ± is plus. Add -60 to 40.

$$x=-\frac{20}{-2}$$

Divide -20 by -2.

$$x=10$$

Now solve the equation x=\frac{-60±40}{-2} when ± is minus. Subtract 40 from -60.

$$x=-\frac{100}{-2}$$

Divide -100 by -2.

$$x=50$$

The equation is now solved.

$$x=10 x=50$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$60x-x^{2}=500$$

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.

$$-x^{2}+60x=500$$

Divide both sides by -1.

$$\frac{-x^{2}+60x}{-1}=\frac{500}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{60}{-1}x=\frac{500}{-1}$$

Divide 60 by -1.

$$x^{2}-60x=\frac{500}{-1}$$

Divide 500 by -1.

$$x^{2}-60x=-500$$

Divide -60, the coefficient of the x term, by 2 to get -30. Then add the square of -30 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-60x+\left(-30\right)^{2}=-500+\left(-30\right)^{2}$$

Square -30.

$$x^{2}-60x+900=-500+900$$

Add -500 to 900.

$$x^{2}-60x+900=400$$

Factor x^{2}-60x+900. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-30\right)^{2}=400$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-30\right)^{2}}=\sqrt{400}$$

Simplify.

$$x-30=20 x-30=-20$$

Add 30 to both sides of the equation.

$$x=50 x=10$$