What are the solutions to the equation roots s^2+6s+13 ?

The solutions to the equation roots s^2+6s+13 are s=-3+2i,s=-3-2i

Find the zeros of roots s^2+6s+13

The zeros of roots s^2+6s+13 are s=-3+2i,s=-3-2i

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roots

s^{2} + 6 s + 13

s = - 3 + 2 i, s = - 3 - 2 i

Solution steps

s^{2} + 6 s + 13

The roots are the intercepts with the x-axis

(y = 0)

s^{2} + 6 s + 13 = 0

Solve with the quadratic formula

s_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 1 \cdot 13}{2 \cdot 1}

Simplify

6^{2} - 4 \cdot 1 \cdot 13 : 4 i

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

Separate the solutions

s_{1} = \frac{- 6 + 4 i}{2 \cdot 1}, s_{2} = \frac{- 6 - 4 i}{2 \cdot 1}

s = \frac{- 6 + 4 i}{2 \cdot 1} : - 3 + 2 i

s = \frac{- 6 - 4 i}{2 \cdot 1} : - 3 - 2 i

The solutions to the quadratic equation are:

s = - 3 + 2 i, s = - 3 - 2 i

2 v^{2} + 4 v = - 10

2 q^{2} + 2 q - 1 = 0

25 v^{2} - 9 = 0

- 2 x + 2 x^{2} = - 3

2 b^{2} + 12 b + 23 = 6

What are the solutions to the equation roots s^2+6s+13 ?
The solutions to the equation roots s^2+6s+13 are s=-3+2i,s=-3-2i
Find the zeros of roots s^2+6s+13
The zeros of roots s^2+6s+13 are s=-3+2i,s=-3-2i