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a+b=7 ab=-60
To solve the equation, factor y^{2}+7y-60 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.
-1,60 -2,30 -3,20 -4,15 -5,12 -6,10
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 -60.
-1+60=59 -2+30=28 -3+20=17 -4+15=11 -5+12=7 -6+10=4
Calculate the sum for each pair.
a=-5 b=12
The solution is the pair that gives sum 7.
\left(y-5\right)\left(y+12\right)
Rewrite factored expression \left(y+a\right)\left(y+b\right) using the obtained values.
y=5 y=-12
To find equation solutions, solve y-5=0 and y+12=0.
a+b=7 ab=1\left(-60\right)=-60
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by-60. To find a and b, set up a system to be solved.
-1,60 -2,30 -3,20 -4,15 -5,12 -6,10
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 -60.
-1+60=59 -2+30=28 -3+20=17 -4+15=11 -5+12=7 -6+10=4
Calculate the sum for each pair.
a=-5 b=12
The solution is the pair that gives sum 7.
\left(y^{2}-5y\right)+\left(12y-60\right)
Rewrite y^{2}+7y-60 as \left(y^{2}-5y\right)+\left(12y-60\right).
y\left(y-5\right)+12\left(y-5\right)
Factor out y in the first and 12 in the second group.
\left(y-5\right)\left(y+12\right)
Factor out common term y-5 by using distributive property.
y=5 y=-12
To find equation solutions, solve y-5=0 and y+12=0.
y^{2}+7y-60=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{-7±\sqrt{7^{2}-4\left(-60\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 7 for b, and -60 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{-7±\sqrt{49-4\left(-60\right)}}{2}
Square 7.
y=\frac{-7±\sqrt{49+240}}{2}
Multiply -4 times -60.
y=\frac{-7±\sqrt{289}}{2}
Add 49 to 240.
y=\frac{-7±17}{2}
Take the square root of 289.
y=\frac{10}{2}
Now solve the equation y=\frac{-7±17}{2} when ± is plus. Add -7 to 17.
y=5
Divide 10 by 2.
y=-\frac{24}{2}
Now solve the equation y=\frac{-7±17}{2} when ± is minus. Subtract 17 from -7.
y=-12
Divide -24 by 2.
y=5 y=-12
The equation is now solved.
y^{2}+7y-60=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.
y^{2}+7y-60-\left(-60\right)=-\left(-60\right)
Add 60 to both sides of the equation.
y^{2}+7y=-\left(-60\right)
Subtracting -60 from itself leaves 0.
y^{2}+7y=60
Subtract -60 from 0.
y^{2}+7y+\left(\frac{7}{2}\right)^{2}=60+\left(\frac{7}{2}\right)^{2}
Divide 7, the coefficient of the x term, by 2 to get \frac{7}{2}. Then add the square of \frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
y^{2}+7y+\frac{49}{4}=60+\frac{49}{4}
Square \frac{7}{2} by squaring both the numerator and the denominator of the fraction.
y^{2}+7y+\frac{49}{4}=\frac{289}{4}
Add 60 to \frac{49}{4}.
\left(y+\frac{7}{2}\right)^{2}=\frac{289}{4}
Factor y^{2}+7y+\frac{49}{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{7}{2}\right)^{2}}=\sqrt{\frac{289}{4}}
Take the square root of both sides of the equation.
y+\frac{7}{2}=\frac{17}{2} y+\frac{7}{2}=-\frac{17}{2}
Simplify.
y=5 y=-12
Subtract \frac{7}{2} from both sides of the equation.
x ^ 2 +7x -60 = 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.
r + s = -7 rs = -60
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{7}{2} - u s = -\frac{7}{2} + u
Two numbers r and s sum up to -7 exactly when the average of the two numbers is \frac{1}{2}*-7 = -\frac{7}{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{7}{2} - u) (-\frac{7}{2} + u) = -60
To solve for unknown quantity u, substitute these in the product equation rs = -60
\frac{49}{4} - u^2 = -60
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -60-\frac{49}{4} = -\frac{289}{4}
Simplify the expression by subtracting \frac{49}{4} on both sides
u^2 = \frac{289}{4} u = \pm\sqrt{\frac{289}{4}} = \pm \frac{17}{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{7}{2} - \frac{17}{2} = -12 s = -\frac{7}{2} + \frac{17}{2} = 5
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.