Solve x^2+5x^=500

$${ x }^{ 2 } +5 { x }^{ } = 500$$

Solve for x

$$x=-25$$

$$x=20$$

Steps Using Factoring

Calculate x to the power of 1 and get x.

$$x^{2}+5x=500$$

Subtract 500 from both sides.

$$x^{2}+5x-500=0$$

To solve the equation, factor x^{2}+5x-500 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=5 ab=-500$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. 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.

$$-1+500=499 -2+250=248 -4+125=121 -5+100=95 -10+50=40 -20+25=5$$

The solution is the pair that gives sum 5.

$$a=-20 b=25$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-20\right)\left(x+25\right)$$

To find equation solutions, solve x-20=0 and x+25=0.

$$x=20 x=-25$$

Steps Using Factoring By Grouping

Calculate x to the power of 1 and get x.

$$x^{2}+5x=500$$

Subtract 500 from both sides.

$$x^{2}+5x-500=0$$

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.

$$a+b=5 ab=1\left(-500\right)=-500$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. 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.

$$-1+500=499 -2+250=248 -4+125=121 -5+100=95 -10+50=40 -20+25=5$$

The solution is the pair that gives sum 5.

$$a=-20 b=25$$

Rewrite x^{2}+5x-500 as \left(x^{2}-20x\right)+\left(25x-500\right).

$$\left(x^{2}-20x\right)+\left(25x-500\right)$$

Factor out x in the first and 25 in the second group.

$$x\left(x-20\right)+25\left(x-20\right)$$

Factor out common term x-20 by using distributive property.

$$\left(x-20\right)\left(x+25\right)$$

To find equation solutions, solve x-20=0 and x+25=0.

$$x=20 x=-25$$

Steps Using the Quadratic Formula

Calculate x to the power of 1 and get x.

$$x^{2}+5x=500$$

Subtract 500 from both sides.

$$x^{2}+5x-500=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 5 for b, and -500 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-500\right)}}{2}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-500\right)}}{2}$$

Multiply -4 times -500.

$$x=\frac{-5±\sqrt{25+2000}}{2}$$

Add 25 to 2000.

$$x=\frac{-5±\sqrt{2025}}{2}$$

Take the square root of 2025.

$$x=\frac{-5±45}{2}$$

Now solve the equation x=\frac{-5±45}{2} when ± is plus. Add -5 to 45.

$$x=\frac{40}{2}$$

Divide 40 by 2.

$$x=20$$

Now solve the equation x=\frac{-5±45}{2} when ± is minus. Subtract 45 from -5.

$$x=-\frac{50}{2}$$

Divide -50 by 2.

$$x=-25$$

The equation is now solved.

$$x=20 x=-25$$

Steps for Completing the Square

Calculate x to the power of 1 and get x.

$$x^{2}+5x=500$$

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.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=500+\left(\frac{5}{2}\right)^{2}$$

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

$$x^{2}+5x+\frac{25}{4}=500+\frac{25}{4}$$

Add 500 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{2025}{4}$$

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}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{2025}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{2025}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{45}{2} x+\frac{5}{2}=-\frac{45}{2}$$

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

$$x=20 x=-25$$