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

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

Find the zeros of roots s^2+4s+13

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

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roots

s^{2} + 4 s + 13

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

Solution steps

s^{2} + 4 s + 13

The roots are the intercepts with the x-axis

(y = 0)

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

4 x^{2} + 11 x - 108 = 0

y, x^{4} y - 8 x y + 3 x y^{2} = 9

x^{2} = - 147

x^{2} = - 166

- 6 x^{2} + 24 x - 120 = 0

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