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14n^{2}-118n+14=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.
n=\frac{-\left(-118\right)±\sqrt{\left(-118\right)^{2}-4\times 14\times 14}}{2\times 14}
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.
n=\frac{-\left(-118\right)±\sqrt{13924-4\times 14\times 14}}{2\times 14}
Square -118.
n=\frac{-\left(-118\right)±\sqrt{13924-56\times 14}}{2\times 14}
Multiply -4 times 14.
n=\frac{-\left(-118\right)±\sqrt{13924-784}}{2\times 14}
Multiply -56 times 14.
n=\frac{-\left(-118\right)±\sqrt{13140}}{2\times 14}
Add 13924 to -784.
n=\frac{-\left(-118\right)±6\sqrt{365}}{2\times 14}
Take the square root of 13140.
n=\frac{118±6\sqrt{365}}{2\times 14}
The opposite of -118 is 118.
n=\frac{118±6\sqrt{365}}{28}
Multiply 2 times 14.
n=\frac{6\sqrt{365}+118}{28}
Now solve the equation n=\frac{118±6\sqrt{365}}{28} when ± is plus. Add 118 to 6\sqrt{365}.
n=\frac{3\sqrt{365}+59}{14}
Divide 118+6\sqrt{365} by 28.
n=\frac{118-6\sqrt{365}}{28}
Now solve the equation n=\frac{118±6\sqrt{365}}{28} when ± is minus. Subtract 6\sqrt{365} from 118.
n=\frac{59-3\sqrt{365}}{14}
Divide 118-6\sqrt{365} by 28.
14n^{2}-118n+14=14\left(n-\frac{3\sqrt{365}+59}{14}\right)\left(n-\frac{59-3\sqrt{365}}{14}\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{59+3\sqrt{365}}{14} for x_{1} and \frac{59-3\sqrt{365}}{14} for x_{2}.
x ^ 2 -\frac{59}{7}x +1 = 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 14
r + s = \frac{59}{7} rs = 1
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{59}{14} - u s = \frac{59}{14} + u
Two numbers r and s sum up to \frac{59}{7} exactly when the average of the two numbers is \frac{1}{2}*\frac{59}{7} = \frac{59}{14}. 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{59}{14} - u) (\frac{59}{14} + u) = 1
To solve for unknown quantity u, substitute these in the product equation rs = 1
\frac{3481}{196} - u^2 = 1
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 1-\frac{3481}{196} = -\frac{3285}{196}
Simplify the expression by subtracting \frac{3481}{196} on both sides
u^2 = \frac{3285}{196} u = \pm\sqrt{\frac{3285}{196}} = \pm \frac{\sqrt{3285}}{14}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{59}{14} - \frac{\sqrt{3285}}{14} = 0.120 s = \frac{59}{14} + \frac{\sqrt{3285}}{14} = 8.308
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.