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2\left(-x^{2}-x+20\right)
Factor out 2.
a+b=-1 ab=-20=-20
Consider -x^{2}-x+20. Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx+20. To find a and b, set up a system to be solved.
1,-20 2,-10 4,-5
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 -20.
1-20=-19 2-10=-8 4-5=-1
Calculate the sum for each pair.
a=4 b=-5
The solution is the pair that gives sum -1.
\left(-x^{2}+4x\right)+\left(-5x+20\right)
Rewrite -x^{2}-x+20 as \left(-x^{2}+4x\right)+\left(-5x+20\right).
x\left(-x+4\right)+5\left(-x+4\right)
Factor out x in the first and 5 in the second group.
\left(-x+4\right)\left(x+5\right)
Factor out common term -x+4 by using distributive property.
2\left(-x+4\right)\left(x+5\right)
Rewrite the complete factored expression.
-2x^{2}-2x+40=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(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-2\right)\times 40}}{2\left(-2\right)}
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(-2\right)±\sqrt{4-4\left(-2\right)\times 40}}{2\left(-2\right)}
Square -2.
x=\frac{-\left(-2\right)±\sqrt{4+8\times 40}}{2\left(-2\right)}
Multiply -4 times -2.
x=\frac{-\left(-2\right)±\sqrt{4+320}}{2\left(-2\right)}
Multiply 8 times 40.
x=\frac{-\left(-2\right)±\sqrt{324}}{2\left(-2\right)}
Add 4 to 320.
x=\frac{-\left(-2\right)±18}{2\left(-2\right)}
Take the square root of 324.
x=\frac{2±18}{2\left(-2\right)}
The opposite of -2 is 2.
x=\frac{2±18}{-4}
Multiply 2 times -2.
x=\frac{20}{-4}
Now solve the equation x=\frac{2±18}{-4} when ± is plus. Add 2 to 18.
x=-5
Divide 20 by -4.
x=-\frac{16}{-4}
Now solve the equation x=\frac{2±18}{-4} when ± is minus. Subtract 18 from 2.
x=4
Divide -16 by -4.
-2x^{2}-2x+40=-2\left(x-\left(-5\right)\right)\left(x-4\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -5 for x_{1} and 4 for x_{2}.
-2x^{2}-2x+40=-2\left(x+5\right)\left(x-4\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
x ^ 2 +1x -20 = 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 = -1 rs = -20
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) = -20
To solve for unknown quantity u, substitute these in the product equation rs = -20
\frac{1}{4} - u^2 = -20
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -20-\frac{1}{4} = -\frac{81}{4}
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = \frac{81}{4} u = \pm\sqrt{\frac{81}{4}} = \pm \frac{9}{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{9}{2} = -5 s = -\frac{1}{2} + \frac{9}{2} = 4
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.