Solve for n
n = \frac{\sqrt{969} + 27}{2} \approx 29.064382416
n=\frac{27-\sqrt{969}}{2}\approx -2.064382416
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n^{2}-27n-60=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{-\left(-27\right)±\sqrt{\left(-27\right)^{2}-4\left(-60\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -27 for b, and -60 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
n=\frac{-\left(-27\right)±\sqrt{729-4\left(-60\right)}}{2}
Square -27.
n=\frac{-\left(-27\right)±\sqrt{729+240}}{2}
Multiply -4 times -60.
n=\frac{-\left(-27\right)±\sqrt{969}}{2}
Add 729 to 240.
n=\frac{27±\sqrt{969}}{2}
The opposite of -27 is 27.
n=\frac{\sqrt{969}+27}{2}
Now solve the equation n=\frac{27±\sqrt{969}}{2} when ± is plus. Add 27 to \sqrt{969}.
n=\frac{27-\sqrt{969}}{2}
Now solve the equation n=\frac{27±\sqrt{969}}{2} when ± is minus. Subtract \sqrt{969} from 27.
n=\frac{\sqrt{969}+27}{2} n=\frac{27-\sqrt{969}}{2}
The equation is now solved.
n^{2}-27n-60=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.
n^{2}-27n-60-\left(-60\right)=-\left(-60\right)
Add 60 to both sides of the equation.
n^{2}-27n=-\left(-60\right)
Subtracting -60 from itself leaves 0.
n^{2}-27n=60
Subtract -60 from 0.
n^{2}-27n+\left(-\frac{27}{2}\right)^{2}=60+\left(-\frac{27}{2}\right)^{2}
Divide -27, the coefficient of the x term, by 2 to get -\frac{27}{2}. Then add the square of -\frac{27}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
n^{2}-27n+\frac{729}{4}=60+\frac{729}{4}
Square -\frac{27}{2} by squaring both the numerator and the denominator of the fraction.
n^{2}-27n+\frac{729}{4}=\frac{969}{4}
Add 60 to \frac{729}{4}.
\left(n-\frac{27}{2}\right)^{2}=\frac{969}{4}
Factor n^{2}-27n+\frac{729}{4}. 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{27}{2}\right)^{2}}=\sqrt{\frac{969}{4}}
Take the square root of both sides of the equation.
n-\frac{27}{2}=\frac{\sqrt{969}}{2} n-\frac{27}{2}=-\frac{\sqrt{969}}{2}
Simplify.
n=\frac{\sqrt{969}+27}{2} n=\frac{27-\sqrt{969}}{2}
Add \frac{27}{2} to both sides of the equation.
x ^ 2 -27x -60 = 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.
r + s = 27 rs = -60
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{27}{2} - u s = \frac{27}{2} + u
Two numbers r and s sum up to 27 exactly when the average of the two numbers is \frac{1}{2}*27 = \frac{27}{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{27}{2} - u) (\frac{27}{2} + u) = -60
To solve for unknown quantity u, substitute these in the product equation rs = -60
\frac{729}{4} - u^2 = -60
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -60-\frac{729}{4} = -\frac{969}{4}
Simplify the expression by subtracting \frac{729}{4} on both sides
u^2 = \frac{969}{4} u = \pm\sqrt{\frac{969}{4}} = \pm \frac{\sqrt{969}}{2}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{27}{2} - \frac{\sqrt{969}}{2} = -2.064 s = \frac{27}{2} + \frac{\sqrt{969}}{2} = 29.064
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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