What are the solutions to the equation n^2+4n-32=0 ?

The solutions to the equation n^2+4n-32=0 are n=4,n=-8

Find the zeros of n^2+4n-32=0

The zeros of n^2+4n-32=0 are n=4,n=-8

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n^{2} + 4 n - 32 = 0

n = 4, n = - 8

Solution steps

n^{2} + 4 n - 32 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 32 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 32) = 12

n_{1, 2} = \frac{- 4 \pm 12}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- 4 + 12}{2 \cdot 1}, n_{2} = \frac{- 4 - 12}{2 \cdot 1}

n = \frac{- 4 + 12}{2 \cdot 1} : 4

n = \frac{- 4 - 12}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

n = 4, n = - 8

- 2 x^{2} + 5 x + 3

5.400447427 x^{2} + 8.145 x - 41.59 = 0

6 x^{2} = - 11 x + 10

21 x^{2} - 42 x = 0

12 - 3 y^{2} = 0

What are the solutions to the equation n^2+4n-32=0 ?
The solutions to the equation n^2+4n-32=0 are n=4,n=-8
Find the zeros of n^2+4n-32=0
The zeros of n^2+4n-32=0 are n=4,n=-8