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a+b=-7 ab=2\left(-30\right)=-60
Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx-30. 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 negative, the negative number has greater absolute value than the positive. 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=-12 b=5
The solution is the pair that gives sum -7.
\left(2x^{2}-12x\right)+\left(5x-30\right)
Rewrite 2x^{2}-7x-30 as \left(2x^{2}-12x\right)+\left(5x-30\right).
2x\left(x-6\right)+5\left(x-6\right)
Factor out 2x in the first and 5 in the second group.
\left(x-6\right)\left(2x+5\right)
Factor out common term x-6 by using distributive property.
2x^{2}-7x-30=0
Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.
x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 2\left(-30\right)}}{2\times 2}
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.
x=\frac{-\left(-7\right)±\sqrt{49-4\times 2\left(-30\right)}}{2\times 2}
Square -7.
x=\frac{-\left(-7\right)±\sqrt{49-8\left(-30\right)}}{2\times 2}
Multiply -4 times 2.
x=\frac{-\left(-7\right)±\sqrt{49+240}}{2\times 2}
Multiply -8 times -30.
x=\frac{-\left(-7\right)±\sqrt{289}}{2\times 2}
Add 49 to 240.
x=\frac{-\left(-7\right)±17}{2\times 2}
Take the square root of 289.
x=\frac{7±17}{2\times 2}
The opposite of -7 is 7.
x=\frac{7±17}{4}
Multiply 2 times 2.
x=\frac{24}{4}
Now solve the equation x=\frac{7±17}{4} when ± is plus. Add 7 to 17.
x=6
Divide 24 by 4.
x=-\frac{10}{4}
Now solve the equation x=\frac{7±17}{4} when ± is minus. Subtract 17 from 7.
x=-\frac{5}{2}
Reduce the fraction \frac{-10}{4} to lowest terms by extracting and canceling out 2.
2x^{2}-7x-30=2\left(x-6\right)\left(x-\left(-\frac{5}{2}\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 6 for x_{1} and -\frac{5}{2} for x_{2}.
2x^{2}-7x-30=2\left(x-6\right)\left(x+\frac{5}{2}\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
2x^{2}-7x-30=2\left(x-6\right)\times \frac{2x+5}{2}
Add \frac{5}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
2x^{2}-7x-30=\left(x-6\right)\left(2x+5\right)
Cancel out 2, the greatest common factor in 2 and 2.
x ^ 2 -\frac{7}{2}x -15 = 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{7}{2} rs = -15
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}{4} - u s = \frac{7}{4} + u
Two numbers r and s sum up to \frac{7}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{7}{2} = \frac{7}{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{7}{4} - u) (\frac{7}{4} + u) = -15
To solve for unknown quantity u, substitute these in the product equation rs = -15
\frac{49}{16} - u^2 = -15
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -15-\frac{49}{16} = -\frac{289}{16}
Simplify the expression by subtracting \frac{49}{16} on both sides
u^2 = \frac{289}{16} u = \pm\sqrt{\frac{289}{16}} = \pm \frac{17}{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{7}{4} - \frac{17}{4} = -2.500 s = \frac{7}{4} + \frac{17}{4} = 6
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.