What are the solutions to the equation s^2+11s+36=0 ?

The solutions to the equation s^2+11s+36=0 are s=-11/2+i(sqrt(23))/2 ,s=-11/2-i(sqrt(23))/2

Find the zeros of s^2+11s+36=0

The zeros of s^2+11s+36=0 are s=-11/2+i(sqrt(23))/2 ,s=-11/2-i(sqrt(23))/2

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s^{2} + 11 s + 36 = 0

s = - \frac{11}{2} + i \frac{23}{2}, s = - \frac{11}{2} - i \frac{23}{2}

Solution steps

s^{2} + 11 s + 36 = 0

Solve with the quadratic formula

s_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 1 \cdot 36}{2 \cdot 1}

Simplify

1 1^{2} - 4 \cdot 1 \cdot 36 : 23 i

s_{1, 2} = \frac{- 11 \pm 23 i}{2 \cdot 1}

Separate the solutions

s_{1} = \frac{- 11 + 23 i}{2 \cdot 1}, s_{2} = \frac{- 11 - 23 i}{2 \cdot 1}

s = \frac{- 11 + 23 i}{2 \cdot 1} : - \frac{11}{2} + i \frac{23}{2}

s = \frac{- 11 - 23 i}{2 \cdot 1} : - \frac{11}{2} - i \frac{23}{2}

The solutions to the quadratic equation are:

s = - \frac{11}{2} + i \frac{23}{2}, s = - \frac{11}{2} - i \frac{23}{2}

(8 + 7 i) (2 + 8 i)

73 x + 144 = 2 x + 71

- \frac{3}{2} - 2

- 5 \leq 2 p + 3 < 13

3 - 2 = 5 - 1

What are the solutions to the equation s^2+11s+36=0 ?
The solutions to the equation s^2+11s+36=0 are s=-11/2+i(sqrt(23))/2 ,s=-11/2-i(sqrt(23))/2
Find the zeros of s^2+11s+36=0
The zeros of s^2+11s+36=0 are s=-11/2+i(sqrt(23))/2 ,s=-11/2-i(sqrt(23))/2