What are the solutions to the equation p^2+5p+36=0 ?

The solutions to the equation p^2+5p+36=0 are p=-5/2+i(sqrt(119))/2 ,p=-5/2-i(sqrt(119))/2

Find the zeros of p^2+5p+36=0

The zeros of p^2+5p+36=0 are p=-5/2+i(sqrt(119))/2 ,p=-5/2-i(sqrt(119))/2

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p^{2} + 5 p + 36 = 0

p = - \frac{5}{2} + i \frac{119}{2}, p = - \frac{5}{2} - i \frac{119}{2}

Solution steps

p^{2} + 5 p + 36 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot 36}{2 \cdot 1}

Simplify

5^{2} - 4 \cdot 1 \cdot 36 : 119 i

p_{1, 2} = \frac{- 5 \pm 119 i}{2 \cdot 1}

Separate the solutions

p_{1} = \frac{- 5 + 119 i}{2 \cdot 1}, p_{2} = \frac{- 5 - 119 i}{2 \cdot 1}

p = \frac{- 5 + 119 i}{2 \cdot 1} : - \frac{5}{2} + i \frac{119}{2}

p = \frac{- 5 - 119 i}{2 \cdot 1} : - \frac{5}{2} - i \frac{119}{2}

The solutions to the quadratic equation are:

p = - \frac{5}{2} + i \frac{119}{2}, p = - \frac{5}{2} - i \frac{119}{2}

n^{3} + 6 n^{2} - 2 n - 12

\frac{8}{5} x - \frac{1}{12} = - 6

8 \cdot 24

6 x^{2} + 10

m^{2} + 5 m = 0

What are the solutions to the equation p^2+5p+36=0 ?
The solutions to the equation p^2+5p+36=0 are p=-5/2+i(sqrt(119))/2 ,p=-5/2-i(sqrt(119))/2
Find the zeros of p^2+5p+36=0
The zeros of p^2+5p+36=0 are p=-5/2+i(sqrt(119))/2 ,p=-5/2-i(sqrt(119))/2