What are the solutions to the equation s^2+12s+27=0 ?

The solutions to the equation s^2+12s+27=0 are s=-3,s=-9

Find the zeros of s^2+12s+27=0

The zeros of s^2+12s+27=0 are s=-3,s=-9

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s^{2} + 12 s + 27 = 0

s = - 3, s = - 9

Solution steps

s^{2} + 12 s + 27 = 0

Solve with the quadratic formula

s_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot 27}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot 27 = 6

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

Separate the solutions

s_{1} = \frac{- 12 + 6}{2 \cdot 1}, s_{2} = \frac{- 12 - 6}{2 \cdot 1}

s = \frac{- 12 + 6}{2 \cdot 1} : - 3

s = \frac{- 12 - 6}{2 \cdot 1} : - 9

The solutions to the quadratic equation are:

s = - 3, s = - 9

(- 12 + 3 i) - (- 2 + i)

5 + 4 - \frac{x}{4} \geq 12

x^{2} - 5 x + 10 = 0

2 x^{2} + 9 x - 518 = 0

(x + 2) (x - 3) < x^{2} + 2

What are the solutions to the equation s^2+12s+27=0 ?
The solutions to the equation s^2+12s+27=0 are s=-3,s=-9
Find the zeros of s^2+12s+27=0
The zeros of s^2+12s+27=0 are s=-3,s=-9