What are the solutions to the equation roots s^2+6s+25 ?

The solutions to the equation roots s^2+6s+25 are s=-3+4i,s=-3-4i

Find the zeros of roots s^2+6s+25

The zeros of roots s^2+6s+25 are s=-3+4i,s=-3-4i

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roots

s^{2} + 6 s + 25

s = - 3 + 4 i, s = - 3 - 4 i

Solution steps

s^{2} + 6 s + 25

The roots are the intercepts with the x-axis

(y = 0)

s^{2} + 6 s + 25 = 0

Solve with the quadratic formula

s_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 1 \cdot 25}{2 \cdot 1}

Simplify

6^{2} - 4 \cdot 1 \cdot 25 : 8 i

s_{1, 2} = \frac{- 6 \pm 8 i}{2 \cdot 1}

Separate the solutions

s_{1} = \frac{- 6 + 8 i}{2 \cdot 1}, s_{2} = \frac{- 6 - 8 i}{2 \cdot 1}

s = \frac{- 6 + 8 i}{2 \cdot 1} : - 3 + 4 i

s = \frac{- 6 - 8 i}{2 \cdot 1} : - 3 - 4 i

The solutions to the quadratic equation are:

s = - 3 + 4 i, s = - 3 - 4 i

35 x^{2} + 17 x = - 2

x^{2} + 18 x + 81 = 0

x^{2} + 18 x - 21 = 0

2 x^{2} - 4 x + 4 = 3 x

9 x^{2} = - 15 x

What are the solutions to the equation roots s^2+6s+25 ?
The solutions to the equation roots s^2+6s+25 are s=-3+4i,s=-3-4i
Find the zeros of roots s^2+6s+25
The zeros of roots s^2+6s+25 are s=-3+4i,s=-3-4i