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a+b=-3 ab=5\left(-36\right)=-180
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5y^{2}+ay+by-36. To find a and b, set up a system to be solved.
1,-180 2,-90 3,-60 4,-45 5,-36 6,-30 9,-20 10,-18 12,-15
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -180.
1-180=-179 2-90=-88 3-60=-57 4-45=-41 5-36=-31 6-30=-24 9-20=-11 10-18=-8 12-15=-3
Calculate the sum for each pair.
a=-15 b=12
The solution is the pair that gives sum -3.
\left(5y^{2}-15y\right)+\left(12y-36\right)
Rewrite 5y^{2}-3y-36 as \left(5y^{2}-15y\right)+\left(12y-36\right).
5y\left(y-3\right)+12\left(y-3\right)
Factor out 5y in the first and 12 in the second group.
\left(y-3\right)\left(5y+12\right)
Factor out common term y-3 by using distributive property.
y=3 y=-\frac{12}{5}
To find equation solutions, solve y-3=0 and 5y+12=0.
5y^{2}-3y-36=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{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 5\left(-36\right)}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -3 for b, and -36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{-\left(-3\right)±\sqrt{9-4\times 5\left(-36\right)}}{2\times 5}
Square -3.
y=\frac{-\left(-3\right)±\sqrt{9-20\left(-36\right)}}{2\times 5}
Multiply -4 times 5.
y=\frac{-\left(-3\right)±\sqrt{9+720}}{2\times 5}
Multiply -20 times -36.
y=\frac{-\left(-3\right)±\sqrt{729}}{2\times 5}
Add 9 to 720.
y=\frac{-\left(-3\right)±27}{2\times 5}
Take the square root of 729.
y=\frac{3±27}{2\times 5}
The opposite of -3 is 3.
y=\frac{3±27}{10}
Multiply 2 times 5.
y=\frac{30}{10}
Now solve the equation y=\frac{3±27}{10} when ± is plus. Add 3 to 27.
y=3
Divide 30 by 10.
y=-\frac{24}{10}
Now solve the equation y=\frac{3±27}{10} when ± is minus. Subtract 27 from 3.
y=-\frac{12}{5}
Reduce the fraction \frac{-24}{10} to lowest terms by extracting and canceling out 2.
y=3 y=-\frac{12}{5}
The equation is now solved.
5y^{2}-3y-36=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.
5y^{2}-3y-36-\left(-36\right)=-\left(-36\right)
Add 36 to both sides of the equation.
5y^{2}-3y=-\left(-36\right)
Subtracting -36 from itself leaves 0.
5y^{2}-3y=36
Subtract -36 from 0.
\frac{5y^{2}-3y}{5}=\frac{36}{5}
Divide both sides by 5.
y^{2}-\frac{3}{5}y=\frac{36}{5}
Dividing by 5 undoes the multiplication by 5.
y^{2}-\frac{3}{5}y+\left(-\frac{3}{10}\right)^{2}=\frac{36}{5}+\left(-\frac{3}{10}\right)^{2}
Divide -\frac{3}{5}, the coefficient of the x term, by 2 to get -\frac{3}{10}. Then add the square of -\frac{3}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
y^{2}-\frac{3}{5}y+\frac{9}{100}=\frac{36}{5}+\frac{9}{100}
Square -\frac{3}{10} by squaring both the numerator and the denominator of the fraction.
y^{2}-\frac{3}{5}y+\frac{9}{100}=\frac{729}{100}
Add \frac{36}{5} to \frac{9}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(y-\frac{3}{10}\right)^{2}=\frac{729}{100}
Factor y^{2}-\frac{3}{5}y+\frac{9}{100}. 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{3}{10}\right)^{2}}=\sqrt{\frac{729}{100}}
Take the square root of both sides of the equation.
y-\frac{3}{10}=\frac{27}{10} y-\frac{3}{10}=-\frac{27}{10}
Simplify.
y=3 y=-\frac{12}{5}
Add \frac{3}{10} to both sides of the equation.
x ^ 2 -\frac{3}{5}x -\frac{36}{5} = 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 5
r + s = \frac{3}{5} rs = -\frac{36}{5}
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{3}{10} - u s = \frac{3}{10} + u
Two numbers r and s sum up to \frac{3}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{3}{5} = \frac{3}{10}. 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{3}{10} - u) (\frac{3}{10} + u) = -\frac{36}{5}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{36}{5}
\frac{9}{100} - u^2 = -\frac{36}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{36}{5}-\frac{9}{100} = -\frac{729}{100}
Simplify the expression by subtracting \frac{9}{100} on both sides
u^2 = \frac{729}{100} u = \pm\sqrt{\frac{729}{100}} = \pm \frac{27}{10}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{3}{10} - \frac{27}{10} = -2.400 s = \frac{3}{10} + \frac{27}{10} = 3
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.