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5x^{2}-62x-7=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(-62\right)±\sqrt{\left(-62\right)^{2}-4\times 5\left(-7\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(-62\right)±\sqrt{3844-4\times 5\left(-7\right)}}{2\times 5}
Square -62.
x=\frac{-\left(-62\right)±\sqrt{3844-20\left(-7\right)}}{2\times 5}
Multiply -4 times 5.
x=\frac{-\left(-62\right)±\sqrt{3844+140}}{2\times 5}
Multiply -20 times -7.
x=\frac{-\left(-62\right)±\sqrt{3984}}{2\times 5}
Add 3844 to 140.
x=\frac{-\left(-62\right)±4\sqrt{249}}{2\times 5}
Take the square root of 3984.
x=\frac{62±4\sqrt{249}}{2\times 5}
The opposite of -62 is 62.
x=\frac{62±4\sqrt{249}}{10}
Multiply 2 times 5.
x=\frac{4\sqrt{249}+62}{10}
Now solve the equation x=\frac{62±4\sqrt{249}}{10} when ± is plus. Add 62 to 4\sqrt{249}.
x=\frac{2\sqrt{249}+31}{5}
Divide 62+4\sqrt{249} by 10.
x=\frac{62-4\sqrt{249}}{10}
Now solve the equation x=\frac{62±4\sqrt{249}}{10} when ± is minus. Subtract 4\sqrt{249} from 62.
x=\frac{31-2\sqrt{249}}{5}
Divide 62-4\sqrt{249} by 10.
5x^{2}-62x-7=5\left(x-\frac{2\sqrt{249}+31}{5}\right)\left(x-\frac{31-2\sqrt{249}}{5}\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{31+2\sqrt{249}}{5} for x_{1} and \frac{31-2\sqrt{249}}{5} for x_{2}.
x ^ 2 -\frac{62}{5}x -\frac{7}{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 = \frac{62}{5} rs = -\frac{7}{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{31}{5} - u s = \frac{31}{5} + u
Two numbers r and s sum up to \frac{62}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{62}{5} = \frac{31}{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>
(\frac{31}{5} - u) (\frac{31}{5} + u) = -\frac{7}{5}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{7}{5}
\frac{961}{25} - u^2 = -\frac{7}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{7}{5}-\frac{961}{25} = -\frac{996}{25}
Simplify the expression by subtracting \frac{961}{25} on both sides
u^2 = \frac{996}{25} u = \pm\sqrt{\frac{996}{25}} = \pm \frac{\sqrt{996}}{5}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{31}{5} - \frac{\sqrt{996}}{5} = -0.112 s = \frac{31}{5} + \frac{\sqrt{996}}{5} = 12.512
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.