What are the solutions to the equation n^2+4n-31=0 ?

The solutions to the equation n^2+4n-31=0 are n=-2+sqrt(35),n=-2-sqrt(35)

Find the zeros of n^2+4n-31=0

The zeros of n^2+4n-31=0 are n=-2+sqrt(35),n=-2-sqrt(35)

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n^{2} + 4 n - 31 = 0

n = - 2 + 35, n = - 2 - 35

Solution steps

n^{2} + 4 n - 31 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 31 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 31) = 235

n_{1, 2} = \frac{- 4 \pm 2 35}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- 4 + 2 35}{2 \cdot 1}, n_{2} = \frac{- 4 - 2 35}{2 \cdot 1}

n = \frac{- 4 + 2 35}{2 \cdot 1} : - 2 + 35

n = \frac{- 4 - 2 35}{2 \cdot 1} : - 2 - 35

The solutions to the quadratic equation are:

n = - 2 + 35, n = - 2 - 35

343 m^{3} + 64 n^{3}

x^{2} - 8 x - 12 = 36

(- 3 a b^{2})^{3}

\frac{2 + 5 i}{5 - 2 i}

x^{3} - 8 x + 8

What are the solutions to the equation n^2+4n-31=0 ?
The solutions to the equation n^2+4n-31=0 are n=-2+sqrt(35),n=-2-sqrt(35)
Find the zeros of n^2+4n-31=0
The zeros of n^2+4n-31=0 are n=-2+sqrt(35),n=-2-sqrt(35)