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2\left(d^{2}-d-30\right)
Factor out 2.
a+b=-1 ab=1\left(-30\right)=-30
Consider d^{2}-d-30. Factor the expression by grouping. First, the expression needs to be rewritten as d^{2}+ad+bd-30. To find a and b, set up a system to be solved.
1,-30 2,-15 3,-10 5,-6
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 -30.
1-30=-29 2-15=-13 3-10=-7 5-6=-1
Calculate the sum for each pair.
a=-6 b=5
The solution is the pair that gives sum -1.
\left(d^{2}-6d\right)+\left(5d-30\right)
Rewrite d^{2}-d-30 as \left(d^{2}-6d\right)+\left(5d-30\right).
d\left(d-6\right)+5\left(d-6\right)
Factor out d in the first and 5 in the second group.
\left(d-6\right)\left(d+5\right)
Factor out common term d-6 by using distributive property.
2\left(d-6\right)\left(d+5\right)
Rewrite the complete factored expression.
2d^{2}-2d-60=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.
d=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 2\left(-60\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.
d=\frac{-\left(-2\right)±\sqrt{4-4\times 2\left(-60\right)}}{2\times 2}
Square -2.
d=\frac{-\left(-2\right)±\sqrt{4-8\left(-60\right)}}{2\times 2}
Multiply -4 times 2.
d=\frac{-\left(-2\right)±\sqrt{4+480}}{2\times 2}
Multiply -8 times -60.
d=\frac{-\left(-2\right)±\sqrt{484}}{2\times 2}
Add 4 to 480.
d=\frac{-\left(-2\right)±22}{2\times 2}
Take the square root of 484.
d=\frac{2±22}{2\times 2}
The opposite of -2 is 2.
d=\frac{2±22}{4}
Multiply 2 times 2.
d=\frac{24}{4}
Now solve the equation d=\frac{2±22}{4} when ± is plus. Add 2 to 22.
d=6
Divide 24 by 4.
d=-\frac{20}{4}
Now solve the equation d=\frac{2±22}{4} when ± is minus. Subtract 22 from 2.
d=-5
Divide -20 by 4.
2d^{2}-2d-60=2\left(d-6\right)\left(d-\left(-5\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 -5 for x_{2}.
2d^{2}-2d-60=2\left(d-6\right)\left(d+5\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
x ^ 2 -1x -30 = 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 = 1 rs = -30
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{1}{2} - u s = \frac{1}{2} + u
Two numbers r and s sum up to 1 exactly when the average of the two numbers is \frac{1}{2}*1 = \frac{1}{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{1}{2} - u) (\frac{1}{2} + u) = -30
To solve for unknown quantity u, substitute these in the product equation rs = -30
\frac{1}{4} - u^2 = -30
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -30-\frac{1}{4} = -\frac{121}{4}
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = \frac{121}{4} u = \pm\sqrt{\frac{121}{4}} = \pm \frac{11}{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{1}{2} - \frac{11}{2} = -5 s = \frac{1}{2} + \frac{11}{2} = 6
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.