What are the solutions to the equation s^2+4s+16=0 ?

The solutions to the equation s^2+4s+16=0 are s=-2+2sqrt(3)i,s=-2-2sqrt(3)i

Find the zeros of s^2+4s+16=0

The zeros of s^2+4s+16=0 are s=-2+2sqrt(3)i,s=-2-2sqrt(3)i

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s^{2} + 4 s + 16 = 0

s = - 2 + 23 i, s = - 2 - 23 i

Solution steps

s^{2} + 4 s + 16 = 0

Solve with the quadratic formula

s_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot 16}{2 \cdot 1}

Simplify

4^{2} - 4 \cdot 1 \cdot 16 : 43 i

s_{1, 2} = \frac{- 4 \pm 4 3 i}{2 \cdot 1}

Separate the solutions

s_{1} = \frac{- 4 + 4 3 i}{2 \cdot 1}, s_{2} = \frac{- 4 - 4 3 i}{2 \cdot 1}

s = \frac{- 4 + 4 3 i}{2 \cdot 1} : - 2 + 23 i

s = \frac{- 4 - 4 3 i}{2 \cdot 1} : - 2 - 23 i

The solutions to the quadratic equation are:

s = - 2 + 23 i, s = - 2 - 23 i

a^{\frac{1}{2}} \cdot a^{\frac{3}{2}}

5 (x + 3) - 6 x \leq 3 (2 x + 1) - 4 x

∣ 2 x + 5 ∣ = 7

\frac{( x + h ) ^{3} - x ^{3}}{h}

- \frac{5}{12} + \frac{4}{9} - \frac{7}{18}

What are the solutions to the equation s^2+4s+16=0 ?
The solutions to the equation s^2+4s+16=0 are s=-2+2sqrt(3)i,s=-2-2sqrt(3)i
Find the zeros of s^2+4s+16=0
The zeros of s^2+4s+16=0 are s=-2+2sqrt(3)i,s=-2-2sqrt(3)i