What are the solutions to the equation s^2+6s+12=0 ?

The solutions to the equation s^2+6s+12=0 are s=-3+sqrt(3)i,s=-3-sqrt(3)i

Find the zeros of s^2+6s+12=0

The zeros of s^2+6s+12=0 are s=-3+sqrt(3)i,s=-3-sqrt(3)i

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s^{2} + 6 s + 12 = 0

s = - 3 + 3 i, s = - 3 - 3 i

Solution steps

s^{2} + 6 s + 12 = 0

Solve with the quadratic formula

s_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

Simplify

6^{2} - 4 \cdot 1 \cdot 12 : 23 i

s_{1, 2} = \frac{- 6 \pm 2 3 i}{2 \cdot 1}

Separate the solutions

s_{1} = \frac{- 6 + 2 3 i}{2 \cdot 1}, s_{2} = \frac{- 6 - 2 3 i}{2 \cdot 1}

s = \frac{- 6 + 2 3 i}{2 \cdot 1} : - 3 + 3 i

s = \frac{- 6 - 2 3 i}{2 \cdot 1} : - 3 - 3 i

The solutions to the quadratic equation are:

s = - 3 + 3 i, s = - 3 - 3 i

- 5 x^{2} + 95 x + 100 = 0

a x^{2} = 4 x - x^{2}

x^{2} - n^{2} x + 2 n = 0

4 x^{2} - 2 x + 5 = 0

(x - 2)^{2} + (x + 4)^{2} = 1

What are the solutions to the equation s^2+6s+12=0 ?
The solutions to the equation s^2+6s+12=0 are s=-3+sqrt(3)i,s=-3-sqrt(3)i
Find the zeros of s^2+6s+12=0
The zeros of s^2+6s+12=0 are s=-3+sqrt(3)i,s=-3-sqrt(3)i