What are the solutions to the equation s^2+4s+12=0 ?

The solutions to the equation s^2+4s+12=0 are s=-2+2sqrt(2)i,s=-2-2sqrt(2)i

Find the zeros of s^2+4s+12=0

The zeros of s^2+4s+12=0 are s=-2+2sqrt(2)i,s=-2-2sqrt(2)i

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s^{2} + 4 s + 12 = 0

s = - 2 + 22 i, s = - 2 - 22 i

Solution steps

s^{2} + 4 s + 12 = 0

Solve with the quadratic formula

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

Simplify

4^{2} - 4 \cdot 1 \cdot 12 : 42 i

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

Separate the solutions

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

s = \frac{- 4 + 4 2 i}{2 \cdot 1} : - 2 + 22 i

s = \frac{- 4 - 4 2 i}{2 \cdot 1} : - 2 - 22 i

The solutions to the quadratic equation are:

s = - 2 + 22 i, s = - 2 - 22 i

2 x^{2} + 7 x = - 1

4 x^{2} - 40 x + 84 = 0

x^{2} + 4 x + (- 1) = 0

x, f = x^{2} + y^{2} + x (y - z) - 4 x - 6 y + 4 z + 30

m^{2} - 4 m - 32 = 0

What are the solutions to the equation s^2+4s+12=0 ?
The solutions to the equation s^2+4s+12=0 are s=-2+2sqrt(2)i,s=-2-2sqrt(2)i
Find the zeros of s^2+4s+12=0
The zeros of s^2+4s+12=0 are s=-2+2sqrt(2)i,s=-2-2sqrt(2)i