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4n^{2}-14n-15=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(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 4\left(-15\right)}}{2\times 4}
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(-14\right)±\sqrt{196-4\times 4\left(-15\right)}}{2\times 4}
Square -14.
n=\frac{-\left(-14\right)±\sqrt{196-16\left(-15\right)}}{2\times 4}
Multiply -4 times 4.
n=\frac{-\left(-14\right)±\sqrt{196+240}}{2\times 4}
Multiply -16 times -15.
n=\frac{-\left(-14\right)±\sqrt{436}}{2\times 4}
Add 196 to 240.
n=\frac{-\left(-14\right)±2\sqrt{109}}{2\times 4}
Take the square root of 436.
n=\frac{14±2\sqrt{109}}{2\times 4}
The opposite of -14 is 14.
n=\frac{14±2\sqrt{109}}{8}
Multiply 2 times 4.
n=\frac{2\sqrt{109}+14}{8}
Now solve the equation n=\frac{14±2\sqrt{109}}{8} when ± is plus. Add 14 to 2\sqrt{109}.
n=\frac{\sqrt{109}+7}{4}
Divide 14+2\sqrt{109} by 8.
n=\frac{14-2\sqrt{109}}{8}
Now solve the equation n=\frac{14±2\sqrt{109}}{8} when ± is minus. Subtract 2\sqrt{109} from 14.
n=\frac{7-\sqrt{109}}{4}
Divide 14-2\sqrt{109} by 8.
4n^{2}-14n-15=4\left(n-\frac{\sqrt{109}+7}{4}\right)\left(n-\frac{7-\sqrt{109}}{4}\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{7+\sqrt{109}}{4} for x_{1} and \frac{7-\sqrt{109}}{4} for x_{2}.
x ^ 2 -\frac{7}{2}x -\frac{15}{4} = 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 4
r + s = \frac{7}{2} rs = -\frac{15}{4}
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{7}{4} - u s = \frac{7}{4} + u
Two numbers r and s sum up to \frac{7}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{7}{2} = \frac{7}{4}. 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{7}{4} - u) (\frac{7}{4} + u) = -\frac{15}{4}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{15}{4}
\frac{49}{16} - u^2 = -\frac{15}{4}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{15}{4}-\frac{49}{16} = -\frac{109}{16}
Simplify the expression by subtracting \frac{49}{16} on both sides
u^2 = \frac{109}{16} u = \pm\sqrt{\frac{109}{16}} = \pm \frac{\sqrt{109}}{4}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{7}{4} - \frac{\sqrt{109}}{4} = -0.860 s = \frac{7}{4} + \frac{\sqrt{109}}{4} = 4.360
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.