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a+b=-3 ab=2\left(-135\right)=-270
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2w^{2}+aw+bw-135. To find a and b, set up a system to be solved.
1,-270 2,-135 3,-90 5,-54 6,-45 9,-30 10,-27 15,-18
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 -270.
1-270=-269 2-135=-133 3-90=-87 5-54=-49 6-45=-39 9-30=-21 10-27=-17 15-18=-3
Calculate the sum for each pair.
a=-18 b=15
The solution is the pair that gives sum -3.
\left(2w^{2}-18w\right)+\left(15w-135\right)
Rewrite 2w^{2}-3w-135 as \left(2w^{2}-18w\right)+\left(15w-135\right).
2w\left(w-9\right)+15\left(w-9\right)
Factor out 2w in the first and 15 in the second group.
\left(w-9\right)\left(2w+15\right)
Factor out common term w-9 by using distributive property.
w=9 w=-\frac{15}{2}
To find equation solutions, solve w-9=0 and 2w+15=0.
2w^{2}-3w-135=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.
w=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2\left(-135\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -3 for b, and -135 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
w=\frac{-\left(-3\right)±\sqrt{9-4\times 2\left(-135\right)}}{2\times 2}
Square -3.
w=\frac{-\left(-3\right)±\sqrt{9-8\left(-135\right)}}{2\times 2}
Multiply -4 times 2.
w=\frac{-\left(-3\right)±\sqrt{9+1080}}{2\times 2}
Multiply -8 times -135.
w=\frac{-\left(-3\right)±\sqrt{1089}}{2\times 2}
Add 9 to 1080.
w=\frac{-\left(-3\right)±33}{2\times 2}
Take the square root of 1089.
w=\frac{3±33}{2\times 2}
The opposite of -3 is 3.
w=\frac{3±33}{4}
Multiply 2 times 2.
w=\frac{36}{4}
Now solve the equation w=\frac{3±33}{4} when ± is plus. Add 3 to 33.
w=9
Divide 36 by 4.
w=-\frac{30}{4}
Now solve the equation w=\frac{3±33}{4} when ± is minus. Subtract 33 from 3.
w=-\frac{15}{2}
Reduce the fraction \frac{-30}{4} to lowest terms by extracting and canceling out 2.
w=9 w=-\frac{15}{2}
The equation is now solved.
2w^{2}-3w-135=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.
2w^{2}-3w-135-\left(-135\right)=-\left(-135\right)
Add 135 to both sides of the equation.
2w^{2}-3w=-\left(-135\right)
Subtracting -135 from itself leaves 0.
2w^{2}-3w=135
Subtract -135 from 0.
\frac{2w^{2}-3w}{2}=\frac{135}{2}
Divide both sides by 2.
w^{2}-\frac{3}{2}w=\frac{135}{2}
Dividing by 2 undoes the multiplication by 2.
w^{2}-\frac{3}{2}w+\left(-\frac{3}{4}\right)^{2}=\frac{135}{2}+\left(-\frac{3}{4}\right)^{2}
Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
w^{2}-\frac{3}{2}w+\frac{9}{16}=\frac{135}{2}+\frac{9}{16}
Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.
w^{2}-\frac{3}{2}w+\frac{9}{16}=\frac{1089}{16}
Add \frac{135}{2} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(w-\frac{3}{4}\right)^{2}=\frac{1089}{16}
Factor w^{2}-\frac{3}{2}w+\frac{9}{16}. 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(w-\frac{3}{4}\right)^{2}}=\sqrt{\frac{1089}{16}}
Take the square root of both sides of the equation.
w-\frac{3}{4}=\frac{33}{4} w-\frac{3}{4}=-\frac{33}{4}
Simplify.
w=9 w=-\frac{15}{2}
Add \frac{3}{4} to both sides of the equation.
x ^ 2 -\frac{3}{2}x -\frac{135}{2} = 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 2
r + s = \frac{3}{2} rs = -\frac{135}{2}
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}{4} - u s = \frac{3}{4} + u
Two numbers r and s sum up to \frac{3}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{3}{2} = \frac{3}{4}. 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}{4} - u) (\frac{3}{4} + u) = -\frac{135}{2}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{135}{2}
\frac{9}{16} - u^2 = -\frac{135}{2}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{135}{2}-\frac{9}{16} = -\frac{1089}{16}
Simplify the expression by subtracting \frac{9}{16} on both sides
u^2 = \frac{1089}{16} u = \pm\sqrt{\frac{1089}{16}} = \pm \frac{33}{4}
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}{4} - \frac{33}{4} = -7.500 s = \frac{3}{4} + \frac{33}{4} = 9
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.