Solve y^2+620y-72000=0

$$y ^ { 2 } + 620 y - 72000 = 0$$

Solve for y

$$y=-720$$

$$y=100$$

Steps Using Factoring

To solve the equation, factor y^{2}+620y-72000 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.

$$a+b=620 ab=-72000$$

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 -72000.

$$-1,72000 -2,36000 -3,24000 -4,18000 -5,14400 -6,12000 -8,9000 -9,8000 -10,7200 -12,6000 -15,4800 -16,4500 -18,4000 -20,3600 -24,3000 -25,2880 -30,2400 -32,2250 -36,2000 -40,1800 -45,1600 -48,1500 -50,1440 -60,1200 -64,1125 -72,1000 -75,960 -80,900 -90,800 -96,750 -100,720 -120,600 -125,576 -144,500 -150,480 -160,450 -180,400 -192,375 -200,360 -225,320 -240,300 -250,288$$

Calculate the sum for each pair.

$$-1+72000=71999 -2+36000=35998 -3+24000=23997 -4+18000=17996 -5+14400=14395 -6+12000=11994 -8+9000=8992 -9+8000=7991 -10+7200=7190 -12+6000=5988 -15+4800=4785 -16+4500=4484 -18+4000=3982 -20+3600=3580 -24+3000=2976 -25+2880=2855 -30+2400=2370 -32+2250=2218 -36+2000=1964 -40+1800=1760 -45+1600=1555 -48+1500=1452 -50+1440=1390 -60+1200=1140 -64+1125=1061 -72+1000=928 -75+960=885 -80+900=820 -90+800=710 -96+750=654 -100+720=620 -120+600=480 -125+576=451 -144+500=356 -150+480=330 -160+450=290 -180+400=220 -192+375=183 -200+360=160 -225+320=95 -240+300=60 -250+288=38$$

The solution is the pair that gives sum 620.

$$a=-100 b=720$$

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

$$\left(y-100\right)\left(y+720\right)$$

To find equation solutions, solve y-100=0 and y+720=0.

$$y=100 y=-720$$

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-72000. To find a and b, set up a system to be solved.

$$a+b=620 ab=1\left(-72000\right)=-72000$$

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 -72000.

$$-1,72000 -2,36000 -3,24000 -4,18000 -5,14400 -6,12000 -8,9000 -9,8000 -10,7200 -12,6000 -15,4800 -16,4500 -18,4000 -20,3600 -24,3000 -25,2880 -30,2400 -32,2250 -36,2000 -40,1800 -45,1600 -48,1500 -50,1440 -60,1200 -64,1125 -72,1000 -75,960 -80,900 -90,800 -96,750 -100,720 -120,600 -125,576 -144,500 -150,480 -160,450 -180,400 -192,375 -200,360 -225,320 -240,300 -250,288$$

Calculate the sum for each pair.

$$-1+72000=71999 -2+36000=35998 -3+24000=23997 -4+18000=17996 -5+14400=14395 -6+12000=11994 -8+9000=8992 -9+8000=7991 -10+7200=7190 -12+6000=5988 -15+4800=4785 -16+4500=4484 -18+4000=3982 -20+3600=3580 -24+3000=2976 -25+2880=2855 -30+2400=2370 -32+2250=2218 -36+2000=1964 -40+1800=1760 -45+1600=1555 -48+1500=1452 -50+1440=1390 -60+1200=1140 -64+1125=1061 -72+1000=928 -75+960=885 -80+900=820 -90+800=710 -96+750=654 -100+720=620 -120+600=480 -125+576=451 -144+500=356 -150+480=330 -160+450=290 -180+400=220 -192+375=183 -200+360=160 -225+320=95 -240+300=60 -250+288=38$$

The solution is the pair that gives sum 620.

$$a=-100 b=720$$

Rewrite y^{2}+620y-72000 as \left(y^{2}-100y\right)+\left(720y-72000\right).

$$\left(y^{2}-100y\right)+\left(720y-72000\right)$$

Factor out y in the first and 720 in the second group.

$$y\left(y-100\right)+720\left(y-100\right)$$

Factor out common term y-100 by using distributive property.

$$\left(y-100\right)\left(y+720\right)$$

To find equation solutions, solve y-100=0 and y+720=0.

$$y=100 y=-720$$

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.

$$y^{2}+620y-72000=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 620 for b, and -72000 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-620±\sqrt{620^{2}-4\left(-72000\right)}}{2}$$

Square 620.

$$y=\frac{-620±\sqrt{384400-4\left(-72000\right)}}{2}$$

Multiply -4 times -72000.

$$y=\frac{-620±\sqrt{384400+288000}}{2}$$

Add 384400 to 288000.

$$y=\frac{-620±\sqrt{672400}}{2}$$

Take the square root of 672400.

$$y=\frac{-620±820}{2}$$

Now solve the equation y=\frac{-620±820}{2} when ± is plus. Add -620 to 820.

$$y=\frac{200}{2}$$

Divide 200 by 2.

$$y=100$$

Now solve the equation y=\frac{-620±820}{2} when ± is minus. Subtract 820 from -620.

$$y=-\frac{1440}{2}$$

Divide -1440 by 2.

$$y=-720$$

The equation is now solved.

$$y=100 y=-720$$

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.

$$y^{2}+620y-72000=0$$

Add 72000 to both sides of the equation.

$$y^{2}+620y-72000-\left(-72000\right)=-\left(-72000\right)$$

Subtracting -72000 from itself leaves 0.

$$y^{2}+620y=-\left(-72000\right)$$

Subtract -72000 from 0.

$$y^{2}+620y=72000$$

Divide 620, the coefficient of the x term, by 2 to get 310. Then add the square of 310 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+620y+310^{2}=72000+310^{2}$$

Square 310.

$$y^{2}+620y+96100=72000+96100$$

Add 72000 to 96100.

$$y^{2}+620y+96100=168100$$

Factor y^{2}+620y+96100. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y+310\right)^{2}=168100$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+310\right)^{2}}=\sqrt{168100}$$

Simplify.

$$y+310=410 y+310=-410$$

Subtract 310 from both sides of the equation.

$$y=100 y=-720$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

$$x ^ 2 +620x -72000 = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = -620 rs = -72000$$

Two numbers r and s sum up to -620 exactly when the average of the two numbers is \frac{1}{2}*-620 = -310. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -310 - u s = -310 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -72000

$$(-310 - u) (-310 + u) = -72000$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$96100 - u^2 = -72000$$

Simplify the expression by subtracting 96100 on both sides

$$-u^2 = -72000-96100 = -168100$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = 168100 u = \pm\sqrt{168100} = \pm 410$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =-310 - 410 = -720 s = -310 + 410 = 100$$