Solve y^2+5y-15000=0

$${ y }^{ 2 } +5y-15000 = 0$$

Solve for y

$$y=-125$$

$$y=120$$

Steps Using Factoring

To solve the equation, factor y^{2}+5y-15000 using formula y^{2}+\left(a+b\right)y+ab=\left(y+a\right)\left(y+b\right). To find a and b, set up a system to be solved.

$$a+b=5 ab=-15000$$

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 -15000.

$$-1,15000 -2,7500 -3,5000 -4,3750 -5,3000 -6,2500 -8,1875 -10,1500 -12,1250 -15,1000 -20,750 -24,625 -25,600 -30,500 -40,375 -50,300 -60,250 -75,200 -100,150 -120,125$$

Calculate the sum for each pair.

$$-1+15000=14999 -2+7500=7498 -3+5000=4997 -4+3750=3746 -5+3000=2995 -6+2500=2494 -8+1875=1867 -10+1500=1490 -12+1250=1238 -15+1000=985 -20+750=730 -24+625=601 -25+600=575 -30+500=470 -40+375=335 -50+300=250 -60+250=190 -75+200=125 -100+150=50 -120+125=5$$

The solution is the pair that gives sum 5.

$$a=-120 b=125$$

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

$$\left(y-120\right)\left(y+125\right)$$

To find equation solutions, solve y-120=0 and y+125=0.

$$y=120 y=-125$$

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 y^{2}+ay+by-15000. To find a and b, set up a system to be solved.

$$a+b=5 ab=1\left(-15000\right)=-15000$$

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 -15000.

$$-1,15000 -2,7500 -3,5000 -4,3750 -5,3000 -6,2500 -8,1875 -10,1500 -12,1250 -15,1000 -20,750 -24,625 -25,600 -30,500 -40,375 -50,300 -60,250 -75,200 -100,150 -120,125$$

Calculate the sum for each pair.

$$-1+15000=14999 -2+7500=7498 -3+5000=4997 -4+3750=3746 -5+3000=2995 -6+2500=2494 -8+1875=1867 -10+1500=1490 -12+1250=1238 -15+1000=985 -20+750=730 -24+625=601 -25+600=575 -30+500=470 -40+375=335 -50+300=250 -60+250=190 -75+200=125 -100+150=50 -120+125=5$$

The solution is the pair that gives sum 5.

$$a=-120 b=125$$

Rewrite y^{2}+5y-15000 as \left(y^{2}-120y\right)+\left(125y-15000\right).

$$\left(y^{2}-120y\right)+\left(125y-15000\right)$$

Factor out y in the first and 125 in the second group.

$$y\left(y-120\right)+125\left(y-120\right)$$

Factor out common term y-120 by using distributive property.

$$\left(y-120\right)\left(y+125\right)$$

To find equation solutions, solve y-120=0 and y+125=0.

$$y=120 y=-125$$

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.

$$y^{2}+5y-15000=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 5 for b, and -15000 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-5±\sqrt{5^{2}-4\left(-15000\right)}}{2}$$

Square 5.

$$y=\frac{-5±\sqrt{25-4\left(-15000\right)}}{2}$$

Multiply -4 times -15000.

$$y=\frac{-5±\sqrt{25+60000}}{2}$$

Add 25 to 60000.

$$y=\frac{-5±\sqrt{60025}}{2}$$

Take the square root of 60025.

$$y=\frac{-5±245}{2}$$

Now solve the equation y=\frac{-5±245}{2} when ± is plus. Add -5 to 245.

$$y=\frac{240}{2}$$

Divide 240 by 2.

$$y=120$$

Now solve the equation y=\frac{-5±245}{2} when ± is minus. Subtract 245 from -5.

$$y=-\frac{250}{2}$$

Divide -250 by 2.

$$y=-125$$

The equation is now solved.

$$y=120 y=-125$$

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.

$$y^{2}+5y-15000=0$$

Add 15000 to both sides of the equation.

$$y^{2}+5y-15000-\left(-15000\right)=-\left(-15000\right)$$

Subtracting -15000 from itself leaves 0.

$$y^{2}+5y=-\left(-15000\right)$$

Subtract -15000 from 0.

$$y^{2}+5y=15000$$

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.

$$y^{2}+5y+\left(\frac{5}{2}\right)^{2}=15000+\left(\frac{5}{2}\right)^{2}$$

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

$$y^{2}+5y+\frac{25}{4}=15000+\frac{25}{4}$$

Add 15000 to \frac{25}{4}.

$$y^{2}+5y+\frac{25}{4}=\frac{60025}{4}$$

Factor y^{2}+5y+\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}.

$$\left(y+\frac{5}{2}\right)^{2}=\frac{60025}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{5}{2}\right)^{2}}=\sqrt{\frac{60025}{4}}$$

Simplify.

$$y+\frac{5}{2}=\frac{245}{2} y+\frac{5}{2}=-\frac{245}{2}$$

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

$$y=120 y=-125$$