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a+b=-27 ab=180
To solve the equation, factor y^{2}-27y+180 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,-180 -2,-90 -3,-60 -4,-45 -5,-36 -6,-30 -9,-20 -10,-18 -12,-15
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 180.
-1-180=-181 -2-90=-92 -3-60=-63 -4-45=-49 -5-36=-41 -6-30=-36 -9-20=-29 -10-18=-28 -12-15=-27
Calculate the sum for each pair.
a=-15 b=-12
The solution is the pair that gives sum -27.
\left(y-15\right)\left(y-12\right)
Rewrite factored expression \left(y+a\right)\left(y+b\right) using the obtained values.
y=15 y=12
To find equation solutions, solve y-15=0 and y-12=0.
a+b=-27 ab=1\times 180=180
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by+180. 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 positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 180.
-1-180=-181 -2-90=-92 -3-60=-63 -4-45=-49 -5-36=-41 -6-30=-36 -9-20=-29 -10-18=-28 -12-15=-27
Calculate the sum for each pair.
a=-15 b=-12
The solution is the pair that gives sum -27.
\left(y^{2}-15y\right)+\left(-12y+180\right)
Rewrite y^{2}-27y+180 as \left(y^{2}-15y\right)+\left(-12y+180\right).
y\left(y-15\right)-12\left(y-15\right)
Factor out y in the first and -12 in the second group.
\left(y-15\right)\left(y-12\right)
Factor out common term y-15 by using distributive property.
y=15 y=12
To find equation solutions, solve y-15=0 and y-12=0.
y^{2}-27y+180=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(-27\right)±\sqrt{\left(-27\right)^{2}-4\times 180}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -27 for b, and 180 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{-\left(-27\right)±\sqrt{729-4\times 180}}{2}
Square -27.
y=\frac{-\left(-27\right)±\sqrt{729-720}}{2}
Multiply -4 times 180.
y=\frac{-\left(-27\right)±\sqrt{9}}{2}
Add 729 to -720.
y=\frac{-\left(-27\right)±3}{2}
Take the square root of 9.
y=\frac{27±3}{2}
The opposite of -27 is 27.
y=\frac{30}{2}
Now solve the equation y=\frac{27±3}{2} when ± is plus. Add 27 to 3.
y=15
Divide 30 by 2.
y=\frac{24}{2}
Now solve the equation y=\frac{27±3}{2} when ± is minus. Subtract 3 from 27.
y=12
Divide 24 by 2.
y=15 y=12
The equation is now solved.
y^{2}-27y+180=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}-27y+180-180=-180
Subtract 180 from both sides of the equation.
y^{2}-27y=-180
Subtracting 180 from itself leaves 0.
y^{2}-27y+\left(-\frac{27}{2}\right)^{2}=-180+\left(-\frac{27}{2}\right)^{2}
Divide -27, the coefficient of the x term, by 2 to get -\frac{27}{2}. Then add the square of -\frac{27}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
y^{2}-27y+\frac{729}{4}=-180+\frac{729}{4}
Square -\frac{27}{2} by squaring both the numerator and the denominator of the fraction.
y^{2}-27y+\frac{729}{4}=\frac{9}{4}
Add -180 to \frac{729}{4}.
\left(y-\frac{27}{2}\right)^{2}=\frac{9}{4}
Factor y^{2}-27y+\frac{729}{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{27}{2}\right)^{2}}=\sqrt{\frac{9}{4}}
Take the square root of both sides of the equation.
y-\frac{27}{2}=\frac{3}{2} y-\frac{27}{2}=-\frac{3}{2}
Simplify.
y=15 y=12
Add \frac{27}{2} to both sides of the equation.
x ^ 2 -27x +180 = 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 = 27 rs = 180
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{27}{2} - u s = \frac{27}{2} + u
Two numbers r and s sum up to 27 exactly when the average of the two numbers is \frac{1}{2}*27 = \frac{27}{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{27}{2} - u) (\frac{27}{2} + u) = 180
To solve for unknown quantity u, substitute these in the product equation rs = 180
\frac{729}{4} - u^2 = 180
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 180-\frac{729}{4} = -\frac{9}{4}
Simplify the expression by subtracting \frac{729}{4} on both sides
u^2 = \frac{9}{4} u = \pm\sqrt{\frac{9}{4}} = \pm \frac{3}{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{27}{2} - \frac{3}{2} = 12 s = \frac{27}{2} + \frac{3}{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.