Solve 25x^2+250x-15000=0

$$25 { x }^{ 2 } +250x-15000=0$$

Solve for x

$$x=-30$$

$$x=20$$

Steps Using Factoring By Grouping

Divide both sides by 25.

$$x^{2}+10x-600=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-600. To find a and b, set up a system to be solved.

$$a+b=10 ab=1\left(-600\right)=-600$$

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 -600.

$$-1,600 -2,300 -3,200 -4,150 -5,120 -6,100 -8,75 -10,60 -12,50 -15,40 -20,30 -24,25$$

Calculate the sum for each pair.

$$-1+600=599 -2+300=298 -3+200=197 -4+150=146 -5+120=115 -6+100=94 -8+75=67 -10+60=50 -12+50=38 -15+40=25 -20+30=10 -24+25=1$$

The solution is the pair that gives sum 10.

$$a=-20 b=30$$

Rewrite x^{2}+10x-600 as \left(x^{2}-20x\right)+\left(30x-600\right).

$$\left(x^{2}-20x\right)+\left(30x-600\right)$$

Factor out x in the first and 30 in the second group.

$$x\left(x-20\right)+30\left(x-20\right)$$

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

$$\left(x-20\right)\left(x+30\right)$$

To find equation solutions, solve x-20=0 and x+30=0.

$$x=20 x=-30$$

Steps Using the Quadratic Formula

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.

$$25x^{2}+250x-15000=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, 250 for b, and -15000 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-250±\sqrt{250^{2}-4\times 25\left(-15000\right)}}{2\times 25}$$

Square 250.

$$x=\frac{-250±\sqrt{62500-4\times 25\left(-15000\right)}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{-250±\sqrt{62500-100\left(-15000\right)}}{2\times 25}$$

Multiply -100 times -15000.

$$x=\frac{-250±\sqrt{62500+1500000}}{2\times 25}$$

Add 62500 to 1500000.

$$x=\frac{-250±\sqrt{1562500}}{2\times 25}$$

Take the square root of 1562500.

$$x=\frac{-250±1250}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{-250±1250}{50}$$

Now solve the equation x=\frac{-250±1250}{50} when ± is plus. Add -250 to 1250.

$$x=\frac{1000}{50}$$

Divide 1000 by 50.

$$x=20$$

Now solve the equation x=\frac{-250±1250}{50} when ± is minus. Subtract 1250 from -250.

$$x=-\frac{1500}{50}$$

Divide -1500 by 50.

$$x=-30$$

The equation is now solved.

$$x=20 x=-30$$

Steps for Completing the Square

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.

$$25x^{2}+250x-15000=0$$

Add 15000 to both sides of the equation.

$$25x^{2}+250x-15000-\left(-15000\right)=-\left(-15000\right)$$

Subtracting -15000 from itself leaves 0.

$$25x^{2}+250x=-\left(-15000\right)$$

Subtract -15000 from 0.

$$25x^{2}+250x=15000$$

Divide both sides by 25.

$$\frac{25x^{2}+250x}{25}=\frac{15000}{25}$$

Dividing by 25 undoes the multiplication by 25.

$$x^{2}+\frac{250}{25}x=\frac{15000}{25}$$

Divide 250 by 25.

$$x^{2}+10x=\frac{15000}{25}$$

Divide 15000 by 25.

$$x^{2}+10x=600$$

Divide 10, the coefficient of the x term, by 2 to get 5. Then add the square of 5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+10x+5^{2}=600+5^{2}$$

Square 5.

$$x^{2}+10x+25=600+25$$

Add 600 to 25.

$$x^{2}+10x+25=625$$

Factor x^{2}+10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+5=25 x+5=-25$$

Subtract 5 from both sides of the equation.

$$x=20 x=-30$$