What are the solutions to the equation 2v^2+4v=-10 ?

The solutions to the equation 2v^2+4v=-10 are v=-1+2i,v=-1-2i

Find the zeros of 2v^2+4v=-10

The zeros of 2v^2+4v=-10 are v=-1+2i,v=-1-2i

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2 v^{2} + 4 v = - 10

v = - 1 + 2 i, v = - 1 - 2 i

Solution steps

2 v^{2} + 4 v = - 10

Move

10

to the left side

2 v^{2} + 4 v + 10 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 2 \cdot 10}{2 \cdot 2}

Simplify

4^{2} - 4 \cdot 2 \cdot 10 : 8 i

v_{1, 2} = \frac{- 4 \pm 8 i}{2 \cdot 2}

Separate the solutions

v_{1} = \frac{- 4 + 8 i}{2 \cdot 2}, v_{2} = \frac{- 4 - 8 i}{2 \cdot 2}

v = \frac{- 4 + 8 i}{2 \cdot 2} : - 1 + 2 i

v = \frac{- 4 - 8 i}{2 \cdot 2} : - 1 - 2 i

The solutions to the quadratic equation are:

v = - 1 + 2 i, v = - 1 - 2 i

2 q^{2} + 2 q - 1 = 0

25 v^{2} - 9 = 0

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

2 b^{2} + 12 b + 23 = 6

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

What are the solutions to the equation 2v^2+4v=-10 ?
The solutions to the equation 2v^2+4v=-10 are v=-1+2i,v=-1-2i
Find the zeros of 2v^2+4v=-10
The zeros of 2v^2+4v=-10 are v=-1+2i,v=-1-2i