What are the solutions to the equation v^2+5v+28=0 ?

The solutions to the equation v^2+5v+28=0 are v=-5/2+i(sqrt(87))/2 ,v=-5/2-i(sqrt(87))/2

Find the zeros of v^2+5v+28=0

The zeros of v^2+5v+28=0 are v=-5/2+i(sqrt(87))/2 ,v=-5/2-i(sqrt(87))/2

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v^{2} + 5 v + 28 = 0

v = - \frac{5}{2} + i \frac{87}{2}, v = - \frac{5}{2} - i \frac{87}{2}

Solution steps

v^{2} + 5 v + 28 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot 28}{2 \cdot 1}

Simplify

5^{2} - 4 \cdot 1 \cdot 28 : 87 i

v_{1, 2} = \frac{- 5 \pm 87 i}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- 5 + 87 i}{2 \cdot 1}, v_{2} = \frac{- 5 - 87 i}{2 \cdot 1}

v = \frac{- 5 + 87 i}{2 \cdot 1} : - \frac{5}{2} + i \frac{87}{2}

v = \frac{- 5 - 87 i}{2 \cdot 1} : - \frac{5}{2} - i \frac{87}{2}

The solutions to the quadratic equation are:

v = - \frac{5}{2} + i \frac{87}{2}, v = - \frac{5}{2} - i \frac{87}{2}

4 u^{2} + 20 u = 0

x^{2} + 13 x - 36 = 0

2 x^{2} - 24 x + 64 = 0

2 x^{2} - 10 x = 12

a^{2} - 4 a + 1 = 0

What are the solutions to the equation v^2+5v+28=0 ?
The solutions to the equation v^2+5v+28=0 are v=-5/2+i(sqrt(87))/2 ,v=-5/2-i(sqrt(87))/2
Find the zeros of v^2+5v+28=0
The zeros of v^2+5v+28=0 are v=-5/2+i(sqrt(87))/2 ,v=-5/2-i(sqrt(87))/2