What are the solutions to the equation s^2+4s+13=0 ?

The solutions to the equation s^2+4s+13=0 are s=-2+3i,s=-2-3i

Find the zeros of s^2+4s+13=0

The zeros of s^2+4s+13=0 are s=-2+3i,s=-2-3i

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s^{2} + 4 s + 13 = 0

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

Solution steps

s^{2} + 4 s + 13 = 0

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

2 x^{2} - x = 15

\frac{2}{3} x + 15 = 17

5 g - 2 \leq 3 g + 8

c - (- 2) \leq 3

x^{2} + 2 x \geq 3

What are the solutions to the equation s^2+4s+13=0 ?
The solutions to the equation s^2+4s+13=0 are s=-2+3i,s=-2-3i
Find the zeros of s^2+4s+13=0
The zeros of s^2+4s+13=0 are s=-2+3i,s=-2-3i