Solve for n
n=\frac{\sqrt{305}-7}{16}\approx 0.654015575
n=\frac{-\sqrt{305}-7}{16}\approx -1.529015575
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8n^{2}+7n-8=0
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{-7±\sqrt{7^{2}-4\times 8\left(-8\right)}}{2\times 8}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 7 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
n=\frac{-7±\sqrt{49-4\times 8\left(-8\right)}}{2\times 8}
Square 7.
n=\frac{-7±\sqrt{49-32\left(-8\right)}}{2\times 8}
Multiply -4 times 8.
n=\frac{-7±\sqrt{49+256}}{2\times 8}
Multiply -32 times -8.
n=\frac{-7±\sqrt{305}}{2\times 8}
Add 49 to 256.
n=\frac{-7±\sqrt{305}}{16}
Multiply 2 times 8.
n=\frac{\sqrt{305}-7}{16}
Now solve the equation n=\frac{-7±\sqrt{305}}{16} when ± is plus. Add -7 to \sqrt{305}.
n=\frac{-\sqrt{305}-7}{16}
Now solve the equation n=\frac{-7±\sqrt{305}}{16} when ± is minus. Subtract \sqrt{305} from -7.
n=\frac{\sqrt{305}-7}{16} n=\frac{-\sqrt{305}-7}{16}
The equation is now solved.
8n^{2}+7n-8=0
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.
8n^{2}+7n-8-\left(-8\right)=-\left(-8\right)
Add 8 to both sides of the equation.
8n^{2}+7n=-\left(-8\right)
Subtracting -8 from itself leaves 0.
8n^{2}+7n=8
Subtract -8 from 0.
\frac{8n^{2}+7n}{8}=\frac{8}{8}
Divide both sides by 8.
n^{2}+\frac{7}{8}n=\frac{8}{8}
Dividing by 8 undoes the multiplication by 8.
n^{2}+\frac{7}{8}n=1
Divide 8 by 8.
n^{2}+\frac{7}{8}n+\left(\frac{7}{16}\right)^{2}=1+\left(\frac{7}{16}\right)^{2}
Divide \frac{7}{8}, the coefficient of the x term, by 2 to get \frac{7}{16}. Then add the square of \frac{7}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
n^{2}+\frac{7}{8}n+\frac{49}{256}=1+\frac{49}{256}
Square \frac{7}{16} by squaring both the numerator and the denominator of the fraction.
n^{2}+\frac{7}{8}n+\frac{49}{256}=\frac{305}{256}
Add 1 to \frac{49}{256}.
\left(n+\frac{7}{16}\right)^{2}=\frac{305}{256}
Factor n^{2}+\frac{7}{8}n+\frac{49}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(n+\frac{7}{16}\right)^{2}}=\sqrt{\frac{305}{256}}
Take the square root of both sides of the equation.
n+\frac{7}{16}=\frac{\sqrt{305}}{16} n+\frac{7}{16}=-\frac{\sqrt{305}}{16}
Simplify.
n=\frac{\sqrt{305}-7}{16} n=\frac{-\sqrt{305}-7}{16}
Subtract \frac{7}{16} from both sides of the equation.
x ^ 2 +\frac{7}{8}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 8
r + s = -\frac{7}{8} 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{7}{16} - u s = -\frac{7}{16} + u
Two numbers r and s sum up to -\frac{7}{8} exactly when the average of the two numbers is \frac{1}{2}*-\frac{7}{8} = -\frac{7}{16}. 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}{16} - u) (-\frac{7}{16} + u) = -1
To solve for unknown quantity u, substitute these in the product equation rs = -1
\frac{49}{256} - u^2 = -1
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -1-\frac{49}{256} = -\frac{305}{256}
Simplify the expression by subtracting \frac{49}{256} on both sides
u^2 = \frac{305}{256} u = \pm\sqrt{\frac{305}{256}} = \pm \frac{\sqrt{305}}{16}
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}{16} - \frac{\sqrt{305}}{16} = -1.529 s = -\frac{7}{16} + \frac{\sqrt{305}}{16} = 0.654
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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