Solve 13x^2+40x-22400=0

$$13 { x }^{ 2 } +40x-22400=0$$

Solve for x

$$x = -\frac{560}{13} = -43\frac{1}{13} \approx -43.076923077$$

$$x=40$$

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 13x^{2}+ax+bx-22400. To find a and b, set up a system to be solved.

$$a+b=40 ab=13\left(-22400\right)=-291200$$

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 -291200.

$$-1,291200 -2,145600 -4,72800 -5,58240 -7,41600 -8,36400 -10,29120 -13,22400 -14,20800 -16,18200 -20,14560 -25,11648 -26,11200 -28,10400 -32,9100 -35,8320 -40,7280 -50,5824 -52,5600 -56,5200 -64,4550 -65,4480 -70,4160 -80,3640 -91,3200 -100,2912 -104,2800 -112,2600 -128,2275 -130,2240 -140,2080 -160,1820 -175,1664 -182,1600 -200,1456 -208,1400 -224,1300 -260,1120 -280,1040 -320,910 -325,896 -350,832 -364,800 -400,728 -416,700 -448,650 -455,640 -520,560$$

Calculate the sum for each pair.

$$-1+291200=291199 -2+145600=145598 -4+72800=72796 -5+58240=58235 -7+41600=41593 -8+36400=36392 -10+29120=29110 -13+22400=22387 -14+20800=20786 -16+18200=18184 -20+14560=14540 -25+11648=11623 -26+11200=11174 -28+10400=10372 -32+9100=9068 -35+8320=8285 -40+7280=7240 -50+5824=5774 -52+5600=5548 -56+5200=5144 -64+4550=4486 -65+4480=4415 -70+4160=4090 -80+3640=3560 -91+3200=3109 -100+2912=2812 -104+2800=2696 -112+2600=2488 -128+2275=2147 -130+2240=2110 -140+2080=1940 -160+1820=1660 -175+1664=1489 -182+1600=1418 -200+1456=1256 -208+1400=1192 -224+1300=1076 -260+1120=860 -280+1040=760 -320+910=590 -325+896=571 -350+832=482 -364+800=436 -400+728=328 -416+700=284 -448+650=202 -455+640=185 -520+560=40$$

The solution is the pair that gives sum 40.

$$a=-520 b=560$$

Rewrite 13x^{2}+40x-22400 as \left(13x^{2}-520x\right)+\left(560x-22400\right).

$$\left(13x^{2}-520x\right)+\left(560x-22400\right)$$

Factor out 13x in the first and 560 in the second group.

$$13x\left(x-40\right)+560\left(x-40\right)$$

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

$$\left(x-40\right)\left(13x+560\right)$$

To find equation solutions, solve x-40=0 and 13x+560=0.

$$x=40 x=-\frac{560}{13}$$

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.

$$13x^{2}+40x-22400=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 13 for a, 40 for b, and -22400 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-40±\sqrt{40^{2}-4\times 13\left(-22400\right)}}{2\times 13}$$

Square 40.

$$x=\frac{-40±\sqrt{1600-4\times 13\left(-22400\right)}}{2\times 13}$$

Multiply -4 times 13.

$$x=\frac{-40±\sqrt{1600-52\left(-22400\right)}}{2\times 13}$$

Multiply -52 times -22400.

$$x=\frac{-40±\sqrt{1600+1164800}}{2\times 13}$$

Add 1600 to 1164800.

$$x=\frac{-40±\sqrt{1166400}}{2\times 13}$$

Take the square root of 1166400.

$$x=\frac{-40±1080}{2\times 13}$$

Multiply 2 times 13.

$$x=\frac{-40±1080}{26}$$

Now solve the equation x=\frac{-40±1080}{26} when ± is plus. Add -40 to 1080.

$$x=\frac{1040}{26}$$

Divide 1040 by 26.

$$x=40$$

Now solve the equation x=\frac{-40±1080}{26} when ± is minus. Subtract 1080 from -40.

$$x=-\frac{1120}{26}$$

Reduce the fraction \frac{-1120}{26} to lowest terms by extracting and canceling out 2.

$$x=-\frac{560}{13}$$

The equation is now solved.

$$x=40 x=-\frac{560}{13}$$

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.

$$13x^{2}+40x-22400=0$$

Add 22400 to both sides of the equation.

$$13x^{2}+40x-22400-\left(-22400\right)=-\left(-22400\right)$$

Subtracting -22400 from itself leaves 0.

$$13x^{2}+40x=-\left(-22400\right)$$

Subtract -22400 from 0.

$$13x^{2}+40x=22400$$

Divide both sides by 13.

$$\frac{13x^{2}+40x}{13}=\frac{22400}{13}$$

Dividing by 13 undoes the multiplication by 13.

$$x^{2}+\frac{40}{13}x=\frac{22400}{13}$$

Divide \frac{40}{13}, the coefficient of the x term, by 2 to get \frac{20}{13}. Then add the square of \frac{20}{13} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{40}{13}x+\left(\frac{20}{13}\right)^{2}=\frac{22400}{13}+\left(\frac{20}{13}\right)^{2}$$

Square \frac{20}{13} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{40}{13}x+\frac{400}{169}=\frac{22400}{13}+\frac{400}{169}$$

Add \frac{22400}{13} to \frac{400}{169} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{40}{13}x+\frac{400}{169}=\frac{291600}{169}$$

Factor x^{2}+\frac{40}{13}x+\frac{400}{169}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{20}{13}\right)^{2}=\frac{291600}{169}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{20}{13}\right)^{2}}=\sqrt{\frac{291600}{169}}$$

Simplify.

$$x+\frac{20}{13}=\frac{540}{13} x+\frac{20}{13}=-\frac{540}{13}$$

Subtract \frac{20}{13} from both sides of the equation.

$$x=40 x=-\frac{560}{13}$$