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5x^{2}-5x-8=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(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 5\left(-8\right)}}{2\times 5}
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(-5\right)±\sqrt{25-4\times 5\left(-8\right)}}{2\times 5}
Square -5.
x=\frac{-\left(-5\right)±\sqrt{25-20\left(-8\right)}}{2\times 5}
Multiply -4 times 5.
x=\frac{-\left(-5\right)±\sqrt{25+160}}{2\times 5}
Multiply -20 times -8.
x=\frac{-\left(-5\right)±\sqrt{185}}{2\times 5}
Add 25 to 160.
x=\frac{5±\sqrt{185}}{2\times 5}
The opposite of -5 is 5.
x=\frac{5±\sqrt{185}}{10}
Multiply 2 times 5.
x=\frac{\sqrt{185}+5}{10}
Now solve the equation x=\frac{5±\sqrt{185}}{10} when ± is plus. Add 5 to \sqrt{185}.
x=\frac{\sqrt{185}}{10}+\frac{1}{2}
Divide 5+\sqrt{185} by 10.
x=\frac{5-\sqrt{185}}{10}
Now solve the equation x=\frac{5±\sqrt{185}}{10} when ± is minus. Subtract \sqrt{185} from 5.
x=-\frac{\sqrt{185}}{10}+\frac{1}{2}
Divide 5-\sqrt{185} by 10.
5x^{2}-5x-8=5\left(x-\left(\frac{\sqrt{185}}{10}+\frac{1}{2}\right)\right)\left(x-\left(-\frac{\sqrt{185}}{10}+\frac{1}{2}\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{2}+\frac{\sqrt{185}}{10} for x_{1} and \frac{1}{2}-\frac{\sqrt{185}}{10} for x_{2}.
x ^ 2 -1x -\frac{8}{5} = 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 5
r + s = 1 rs = -\frac{8}{5}
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) = -\frac{8}{5}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{8}{5}
\frac{1}{4} - u^2 = -\frac{8}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{8}{5}-\frac{1}{4} = -\frac{37}{20}
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = \frac{37}{20} u = \pm\sqrt{\frac{37}{20}} = \pm \frac{\sqrt{37}}{\sqrt{20}}
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{\sqrt{37}}{\sqrt{20}} = -0.860 s = \frac{1}{2} + \frac{\sqrt{37}}{\sqrt{20}} = 1.860
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.