What are the solutions to the equation -v^2+5v+6=-2v^2 ?

The solutions to the equation -v^2+5v+6=-2v^2 are v=-2,v=-3

Find the zeros of -v^2+5v+6=-2v^2

The zeros of -v^2+5v+6=-2v^2 are v=-2,v=-3

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- v^{2} + 5 v + 6 = - 2 v^{2}

v = - 2, v = - 3

Solution steps

- v^{2} + 5 v + 6 = - 2 v^{2}

Move

2 v^{2}

to the left side

v^{2} + 5 v + 6 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot 6}{2 \cdot 1}

5^{2} - 4 \cdot 1 \cdot 6 = 1

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

Separate the solutions

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

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

v = \frac{- 5 - 1}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

v = - 2, v = - 3

9 x^{2} - 9 x - 10 = 0

2 x^{2} - 16 x + 26 = 0

10 m^{2} + 8 m + 3 = 0

- (x + 3)^{2} + 5 = 0

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

What are the solutions to the equation -v^2+5v+6=-2v^2 ?
The solutions to the equation -v^2+5v+6=-2v^2 are v=-2,v=-3
Find the zeros of -v^2+5v+6=-2v^2
The zeros of -v^2+5v+6=-2v^2 are v=-2,v=-3