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7n^{2}+11n-1=2
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.
7n^{2}+11n-1-2=2-2
Subtract 2 from both sides of the equation.
7n^{2}+11n-1-2=0
Subtracting 2 from itself leaves 0.
7n^{2}+11n-3=0
Subtract 2 from -1.
n=\frac{-11±\sqrt{11^{2}-4\times 7\left(-3\right)}}{2\times 7}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 7 for a, 11 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
n=\frac{-11±\sqrt{121-4\times 7\left(-3\right)}}{2\times 7}
Square 11.
n=\frac{-11±\sqrt{121-28\left(-3\right)}}{2\times 7}
Multiply -4 times 7.
n=\frac{-11±\sqrt{121+84}}{2\times 7}
Multiply -28 times -3.
n=\frac{-11±\sqrt{205}}{2\times 7}
Add 121 to 84.
n=\frac{-11±\sqrt{205}}{14}
Multiply 2 times 7.
n=\frac{\sqrt{205}-11}{14}
Now solve the equation n=\frac{-11±\sqrt{205}}{14} when ± is plus. Add -11 to \sqrt{205}.
n=\frac{-\sqrt{205}-11}{14}
Now solve the equation n=\frac{-11±\sqrt{205}}{14} when ± is minus. Subtract \sqrt{205} from -11.
n=\frac{\sqrt{205}-11}{14} n=\frac{-\sqrt{205}-11}{14}
The equation is now solved.
7n^{2}+11n-1=2
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.
7n^{2}+11n-1-\left(-1\right)=2-\left(-1\right)
Add 1 to both sides of the equation.
7n^{2}+11n=2-\left(-1\right)
Subtracting -1 from itself leaves 0.
7n^{2}+11n=3
Subtract -1 from 2.
\frac{7n^{2}+11n}{7}=\frac{3}{7}
Divide both sides by 7.
n^{2}+\frac{11}{7}n=\frac{3}{7}
Dividing by 7 undoes the multiplication by 7.
n^{2}+\frac{11}{7}n+\left(\frac{11}{14}\right)^{2}=\frac{3}{7}+\left(\frac{11}{14}\right)^{2}
Divide \frac{11}{7}, the coefficient of the x term, by 2 to get \frac{11}{14}. Then add the square of \frac{11}{14} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
n^{2}+\frac{11}{7}n+\frac{121}{196}=\frac{3}{7}+\frac{121}{196}
Square \frac{11}{14} by squaring both the numerator and the denominator of the fraction.
n^{2}+\frac{11}{7}n+\frac{121}{196}=\frac{205}{196}
Add \frac{3}{7} to \frac{121}{196} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(n+\frac{11}{14}\right)^{2}=\frac{205}{196}
Factor n^{2}+\frac{11}{7}n+\frac{121}{196}. 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{11}{14}\right)^{2}}=\sqrt{\frac{205}{196}}
Take the square root of both sides of the equation.
n+\frac{11}{14}=\frac{\sqrt{205}}{14} n+\frac{11}{14}=-\frac{\sqrt{205}}{14}
Simplify.
n=\frac{\sqrt{205}-11}{14} n=\frac{-\sqrt{205}-11}{14}
Subtract \frac{11}{14} from both sides of the equation.