Solve x^2+20x-2400=0

$${ x }^{ 2 } +20x-2400=0$$

Solve for x

$$x=-60$$

$$x=40$$

Steps Using Factoring

To solve the equation, factor x^{2}+20x-2400 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.

$$a+b=20 ab=-2400$$

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 -2400.

$$-1,2400 -2,1200 -3,800 -4,600 -5,480 -6,400 -8,300 -10,240 -12,200 -15,160 -16,150 -20,120 -24,100 -25,96 -30,80 -32,75 -40,60 -48,50$$

Calculate the sum for each pair.

$$-1+2400=2399 -2+1200=1198 -3+800=797 -4+600=596 -5+480=475 -6+400=394 -8+300=292 -10+240=230 -12+200=188 -15+160=145 -16+150=134 -20+120=100 -24+100=76 -25+96=71 -30+80=50 -32+75=43 -40+60=20 -48+50=2$$

The solution is the pair that gives sum 20.

$$a=-40 b=60$$

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

$$\left(x-40\right)\left(x+60\right)$$

To find equation solutions, solve x-40=0 and x+60=0.

$$x=40 x=-60$$

Steps Using Factoring By Grouping

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-2400. To find a and b, set up a system to be solved.

$$a+b=20 ab=1\left(-2400\right)=-2400$$

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 -2400.

$$-1,2400 -2,1200 -3,800 -4,600 -5,480 -6,400 -8,300 -10,240 -12,200 -15,160 -16,150 -20,120 -24,100 -25,96 -30,80 -32,75 -40,60 -48,50$$

Calculate the sum for each pair.

$$-1+2400=2399 -2+1200=1198 -3+800=797 -4+600=596 -5+480=475 -6+400=394 -8+300=292 -10+240=230 -12+200=188 -15+160=145 -16+150=134 -20+120=100 -24+100=76 -25+96=71 -30+80=50 -32+75=43 -40+60=20 -48+50=2$$

The solution is the pair that gives sum 20.

$$a=-40 b=60$$

Rewrite x^{2}+20x-2400 as \left(x^{2}-40x\right)+\left(60x-2400\right).

$$\left(x^{2}-40x\right)+\left(60x-2400\right)$$

Factor out x in the first and 60 in the second group.

$$x\left(x-40\right)+60\left(x-40\right)$$

Factor out common term x-40 by using distributive property.

$$\left(x-40\right)\left(x+60\right)$$

To find equation solutions, solve x-40=0 and x+60=0.

$$x=40 x=-60$$

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.

$$x^{2}+20x-2400=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 20 for b, and -2400 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-20±\sqrt{20^{2}-4\left(-2400\right)}}{2}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\left(-2400\right)}}{2}$$

Multiply -4 times -2400.

$$x=\frac{-20±\sqrt{400+9600}}{2}$$

Add 400 to 9600.

$$x=\frac{-20±\sqrt{10000}}{2}$$

Take the square root of 10000.

$$x=\frac{-20±100}{2}$$

Now solve the equation x=\frac{-20±100}{2} when ± is plus. Add -20 to 100.

$$x=\frac{80}{2}$$

Divide 80 by 2.

$$x=40$$

Now solve the equation x=\frac{-20±100}{2} when ± is minus. Subtract 100 from -20.

$$x=-\frac{120}{2}$$

Divide -120 by 2.

$$x=-60$$

The equation is now solved.

$$x=40 x=-60$$

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.

$$x^{2}+20x-2400=0$$

Add 2400 to both sides of the equation.

$$x^{2}+20x-2400-\left(-2400\right)=-\left(-2400\right)$$

Subtracting -2400 from itself leaves 0.

$$x^{2}+20x=-\left(-2400\right)$$

Subtract -2400 from 0.

$$x^{2}+20x=2400$$

Divide 20, the coefficient of the x term, by 2 to get 10. Then add the square of 10 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+20x+10^{2}=2400+10^{2}$$

Square 10.

$$x^{2}+20x+100=2400+100$$

Add 2400 to 100.

$$x^{2}+20x+100=2500$$

Factor x^{2}+20x+100. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+10\right)^{2}=2500$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+10\right)^{2}}=\sqrt{2500}$$

Simplify.

$$x+10=50 x+10=-50$$

Subtract 10 from both sides of the equation.

$$x=40 x=-60$$