What are the solutions to the equation n^2-2n-108=-8 ?

The solutions to the equation n^2-2n-108=-8 are n=1+sqrt(101),n=1-sqrt(101)

Find the zeros of n^2-2n-108=-8

The zeros of n^2-2n-108=-8 are n=1+sqrt(101),n=1-sqrt(101)

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n^{2} - 2 n - 108 = - 8

n = 1 + 101, n = 1 - 101

Decimal Notation

n = 11.04987 \dots, n = - 9.04987 \dots

Solution steps

n^{2} - 2 n - 108 = - 8

Move

8

to the left side

n^{2} - 2 n - 100 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 100 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 100) = 2101

n_{1, 2} = \frac{- ( - 2 ) \pm 2 101}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 2 ) + 2 101}{2 \cdot 1}, n_{2} = \frac{- ( - 2 ) - 2 101}{2 \cdot 1}

n = \frac{- ( - 2 ) + 2 101}{2 \cdot 1} : 1 + 101

n = \frac{- ( - 2 ) - 2 101}{2 \cdot 1} : 1 - 101

The solutions to the quadratic equation are:

n = 1 + 101, n = 1 - 101

(\frac{4}{6})^{4}

\frac{\frac{x}{y} - \frac{y}{x}}{\frac{1}{6 x ^{2}} - \frac{1}{6 y ^{2}}}

0.011 x^{2} + 0.0067 x + 0.2329 = 0.3015

\frac{1}{4} a = \frac{4}{15}

- 6 \leq 3 x < 9

What are the solutions to the equation n^2-2n-108=-8 ?
The solutions to the equation n^2-2n-108=-8 are n=1+sqrt(101),n=1-sqrt(101)
Find the zeros of n^2-2n-108=-8
The zeros of n^2-2n-108=-8 are n=1+sqrt(101),n=1-sqrt(101)