What are the solutions to the equation n^2+4n+72=0 ?

The solutions to the equation n^2+4n+72=0 are n=-2+2sqrt(17)i,n=-2-2sqrt(17)i

Find the zeros of n^2+4n+72=0

The zeros of n^2+4n+72=0 are n=-2+2sqrt(17)i,n=-2-2sqrt(17)i

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n^{2} + 4 n + 72 = 0

n = - 2 + 217 i, n = - 2 - 217 i

Solution steps

n^{2} + 4 n + 72 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot 72}{2 \cdot 1}

Simplify

4^{2} - 4 \cdot 1 \cdot 72 : 417 i

n_{1, 2} = \frac{- 4 \pm 4 17 i}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- 4 + 4 17 i}{2 \cdot 1}, n_{2} = \frac{- 4 - 4 17 i}{2 \cdot 1}

n = \frac{- 4 + 4 17 i}{2 \cdot 1} : - 2 + 217 i

n = \frac{- 4 - 4 17 i}{2 \cdot 1} : - 2 - 217 i

The solutions to the quadratic equation are:

n = - 2 + 217 i, n = - 2 - 217 i

∣ 9 - 2 x ∣ > 5

2 x + 10 < 22

\frac{14}{- 2}

1 + 4 (3 b + 9) - b

8 x^{3} - 16 x^{2} + 4 x

What are the solutions to the equation n^2+4n+72=0 ?
The solutions to the equation n^2+4n+72=0 are n=-2+2sqrt(17)i,n=-2-2sqrt(17)i
Find the zeros of n^2+4n+72=0
The zeros of n^2+4n+72=0 are n=-2+2sqrt(17)i,n=-2-2sqrt(17)i