What are the solutions to the equation n^2+12n+32=0 ?

The solutions to the equation n^2+12n+32=0 are n=-4,n=-8

Find the zeros of n^2+12n+32=0

The zeros of n^2+12n+32=0 are n=-4,n=-8

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n^{2} + 12 n + 32 = 0

n = - 4, n = - 8

Solution steps

n^{2} + 12 n + 32 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot 32}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot 32 = 4

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

n = - 4, n = - 8

3 x^{2} - 13 x = - 10

x, x^{2} = - 4 p y

x^{2} - 9 x = 0

- 3 y^{2} + 2 y + 8 = 0

5232.6 x^{2} - 196250 x - 15072000 = 0

What are the solutions to the equation n^2+12n+32=0 ?
The solutions to the equation n^2+12n+32=0 are n=-4,n=-8
Find the zeros of n^2+12n+32=0
The zeros of n^2+12n+32=0 are n=-4,n=-8