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y^{2}+y-240=0
Divide both sides by 4.
a+b=1 ab=1\left(-240\right)=-240
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by-240. To find a and b, set up a system to be solved.
-1,240 -2,120 -3,80 -4,60 -5,48 -6,40 -8,30 -10,24 -12,20 -15,16
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 -240.
-1+240=239 -2+120=118 -3+80=77 -4+60=56 -5+48=43 -6+40=34 -8+30=22 -10+24=14 -12+20=8 -15+16=1
Calculate the sum for each pair.
a=-15 b=16
The solution is the pair that gives sum 1.
\left(y^{2}-15y\right)+\left(16y-240\right)
Rewrite y^{2}+y-240 as \left(y^{2}-15y\right)+\left(16y-240\right).
y\left(y-15\right)+16\left(y-15\right)
Factor out y in the first and 16 in the second group.
\left(y-15\right)\left(y+16\right)
Factor out common term y-15 by using distributive property.
y=15 y=-16
To find equation solutions, solve y-15=0 and y+16=0.
4y^{2}+4y-960=0
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=\frac{-4±\sqrt{4^{2}-4\times 4\left(-960\right)}}{2\times 4}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 4 for b, and -960 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{-4±\sqrt{16-4\times 4\left(-960\right)}}{2\times 4}
Square 4.
y=\frac{-4±\sqrt{16-16\left(-960\right)}}{2\times 4}
Multiply -4 times 4.
y=\frac{-4±\sqrt{16+15360}}{2\times 4}
Multiply -16 times -960.
y=\frac{-4±\sqrt{15376}}{2\times 4}
Add 16 to 15360.
y=\frac{-4±124}{2\times 4}
Take the square root of 15376.
y=\frac{-4±124}{8}
Multiply 2 times 4.
y=\frac{120}{8}
Now solve the equation y=\frac{-4±124}{8} when ± is plus. Add -4 to 124.
y=15
Divide 120 by 8.
y=-\frac{128}{8}
Now solve the equation y=\frac{-4±124}{8} when ± is minus. Subtract 124 from -4.
y=-16
Divide -128 by 8.
y=15 y=-16
The equation is now solved.
4y^{2}+4y-960=0
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.
4y^{2}+4y-960-\left(-960\right)=-\left(-960\right)
Add 960 to both sides of the equation.
4y^{2}+4y=-\left(-960\right)
Subtracting -960 from itself leaves 0.
4y^{2}+4y=960
Subtract -960 from 0.
\frac{4y^{2}+4y}{4}=\frac{960}{4}
Divide both sides by 4.
y^{2}+\frac{4}{4}y=\frac{960}{4}
Dividing by 4 undoes the multiplication by 4.
y^{2}+y=\frac{960}{4}
Divide 4 by 4.
y^{2}+y=240
Divide 960 by 4.
y^{2}+y+\left(\frac{1}{2}\right)^{2}=240+\left(\frac{1}{2}\right)^{2}
Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
y^{2}+y+\frac{1}{4}=240+\frac{1}{4}
Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.
y^{2}+y+\frac{1}{4}=\frac{961}{4}
Add 240 to \frac{1}{4}.
\left(y+\frac{1}{2}\right)^{2}=\frac{961}{4}
Factor y^{2}+y+\frac{1}{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}.
\sqrt{\left(y+\frac{1}{2}\right)^{2}}=\sqrt{\frac{961}{4}}
Take the square root of both sides of the equation.
y+\frac{1}{2}=\frac{31}{2} y+\frac{1}{2}=-\frac{31}{2}
Simplify.
y=15 y=-16
Subtract \frac{1}{2} from both sides of the equation.
x ^ 2 +1x -240 = 0
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.This is achieved by dividing both sides of the equation by 4
r + s = -1 rs = -240
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 = -\frac{1}{2} - u s = -\frac{1}{2} + u
Two numbers r and s sum up to -1 exactly when the average of the two numbers is \frac{1}{2}*-1 = -\frac{1}{2}. 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>
(-\frac{1}{2} - u) (-\frac{1}{2} + u) = -240
To solve for unknown quantity u, substitute these in the product equation rs = -240
\frac{1}{4} - u^2 = -240
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -240-\frac{1}{4} = -\frac{961}{4}
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = \frac{961}{4} u = \pm\sqrt{\frac{961}{4}} = \pm \frac{31}{2}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{1}{2} - \frac{31}{2} = -16 s = -\frac{1}{2} + \frac{31}{2} = 15
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.