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-4n^{2}+10n+1=8
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.
-4n^{2}+10n+1-8=8-8
Subtract 8 from both sides of the equation.
-4n^{2}+10n+1-8=0
Subtracting 8 from itself leaves 0.
-4n^{2}+10n-7=0
Subtract 8 from 1.
n=\frac{-10±\sqrt{10^{2}-4\left(-4\right)\left(-7\right)}}{2\left(-4\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 10 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
n=\frac{-10±\sqrt{100-4\left(-4\right)\left(-7\right)}}{2\left(-4\right)}
Square 10.
n=\frac{-10±\sqrt{100+16\left(-7\right)}}{2\left(-4\right)}
Multiply -4 times -4.
n=\frac{-10±\sqrt{100-112}}{2\left(-4\right)}
Multiply 16 times -7.
n=\frac{-10±\sqrt{-12}}{2\left(-4\right)}
Add 100 to -112.
n=\frac{-10±2\sqrt{3}i}{2\left(-4\right)}
Take the square root of -12.
n=\frac{-10±2\sqrt{3}i}{-8}
Multiply 2 times -4.
n=\frac{-10+2\sqrt{3}i}{-8}
Now solve the equation n=\frac{-10±2\sqrt{3}i}{-8} when ± is plus. Add -10 to 2i\sqrt{3}.
n=\frac{-\sqrt{3}i+5}{4}
Divide -10+2i\sqrt{3} by -8.
n=\frac{-2\sqrt{3}i-10}{-8}
Now solve the equation n=\frac{-10±2\sqrt{3}i}{-8} when ± is minus. Subtract 2i\sqrt{3} from -10.
n=\frac{5+\sqrt{3}i}{4}
Divide -10-2i\sqrt{3} by -8.
n=\frac{-\sqrt{3}i+5}{4} n=\frac{5+\sqrt{3}i}{4}
The equation is now solved.
-4n^{2}+10n+1=8
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.
-4n^{2}+10n+1-1=8-1
Subtract 1 from both sides of the equation.
-4n^{2}+10n=8-1
Subtracting 1 from itself leaves 0.
-4n^{2}+10n=7
Subtract 1 from 8.
\frac{-4n^{2}+10n}{-4}=\frac{7}{-4}
Divide both sides by -4.
n^{2}+\frac{10}{-4}n=\frac{7}{-4}
Dividing by -4 undoes the multiplication by -4.
n^{2}-\frac{5}{2}n=\frac{7}{-4}
Reduce the fraction \frac{10}{-4} to lowest terms by extracting and canceling out 2.
n^{2}-\frac{5}{2}n=-\frac{7}{4}
Divide 7 by -4.
n^{2}-\frac{5}{2}n+\left(-\frac{5}{4}\right)^{2}=-\frac{7}{4}+\left(-\frac{5}{4}\right)^{2}
Divide -\frac{5}{2}, the coefficient of the x term, by 2 to get -\frac{5}{4}. Then add the square of -\frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
n^{2}-\frac{5}{2}n+\frac{25}{16}=-\frac{7}{4}+\frac{25}{16}
Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.
n^{2}-\frac{5}{2}n+\frac{25}{16}=-\frac{3}{16}
Add -\frac{7}{4} to \frac{25}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(n-\frac{5}{4}\right)^{2}=-\frac{3}{16}
Factor n^{2}-\frac{5}{2}n+\frac{25}{16}. 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{5}{4}\right)^{2}}=\sqrt{-\frac{3}{16}}
Take the square root of both sides of the equation.
n-\frac{5}{4}=\frac{\sqrt{3}i}{4} n-\frac{5}{4}=-\frac{\sqrt{3}i}{4}
Simplify.
n=\frac{5+\sqrt{3}i}{4} n=\frac{-\sqrt{3}i+5}{4}
Add \frac{5}{4} to both sides of the equation.