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-2x^{2}-20x+5=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(-20\right)±\sqrt{\left(-20\right)^{2}-4\left(-2\right)\times 5}}{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(-20\right)±\sqrt{400-4\left(-2\right)\times 5}}{2\left(-2\right)}
Square -20.
x=\frac{-\left(-20\right)±\sqrt{400+8\times 5}}{2\left(-2\right)}
Multiply -4 times -2.
x=\frac{-\left(-20\right)±\sqrt{400+40}}{2\left(-2\right)}
Multiply 8 times 5.
x=\frac{-\left(-20\right)±\sqrt{440}}{2\left(-2\right)}
Add 400 to 40.
x=\frac{-\left(-20\right)±2\sqrt{110}}{2\left(-2\right)}
Take the square root of 440.
x=\frac{20±2\sqrt{110}}{2\left(-2\right)}
The opposite of -20 is 20.
x=\frac{20±2\sqrt{110}}{-4}
Multiply 2 times -2.
x=\frac{2\sqrt{110}+20}{-4}
Now solve the equation x=\frac{20±2\sqrt{110}}{-4} when ± is plus. Add 20 to 2\sqrt{110}.
x=-\frac{\sqrt{110}}{2}-5
Divide 20+2\sqrt{110} by -4.
x=\frac{20-2\sqrt{110}}{-4}
Now solve the equation x=\frac{20±2\sqrt{110}}{-4} when ± is minus. Subtract 2\sqrt{110} from 20.
x=\frac{\sqrt{110}}{2}-5
Divide 20-2\sqrt{110} by -4.
-2x^{2}-20x+5=-2\left(x-\left(-\frac{\sqrt{110}}{2}-5\right)\right)\left(x-\left(\frac{\sqrt{110}}{2}-5\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -5-\frac{\sqrt{110}}{2} for x_{1} and -5+\frac{\sqrt{110}}{2} for x_{2}.
x ^ 2 +10x -\frac{5}{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.
r + s = -10 rs = -\frac{5}{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 = -5 - u s = -5 + u
Two numbers r and s sum up to -10 exactly when the average of the two numbers is \frac{1}{2}*-10 = -5. 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>
(-5 - u) (-5 + u) = -\frac{5}{2}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{2}
25 - u^2 = -\frac{5}{2}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{5}{2}-25 = -\frac{55}{2}
Simplify the expression by subtracting 25 on both sides
u^2 = \frac{55}{2} u = \pm\sqrt{\frac{55}{2}} = \pm \frac{\sqrt{55}}{\sqrt{2}}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-5 - \frac{\sqrt{55}}{\sqrt{2}} = -10.244 s = -5 + \frac{\sqrt{55}}{\sqrt{2}} = 0.244
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.