What are the solutions to the equation v^2+2v-8=0 ?

The solutions to the equation v^2+2v-8=0 are v=2,v=-4

Find the zeros of v^2+2v-8=0

The zeros of v^2+2v-8=0 are v=2,v=-4

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v^{2} + 2 v - 8 = 0

v = 2, v = - 4

Solution steps

v^{2} + 2 v - 8 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 8) = 6

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

Separate the solutions

v_{1} = \frac{- 2 + 6}{2 \cdot 1}, v_{2} = \frac{- 2 - 6}{2 \cdot 1}

v = \frac{- 2 + 6}{2 \cdot 1} : 2

v = \frac{- 2 - 6}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

v = 2, v = - 4

2.5 c - 8 > 0.5

1 + 2 k > 9

\frac{16 x ^{3} y ^{4}}{40 x ^{6} y ^{8}}

(\frac{2}{3})^{\frac{3}{2}}

\frac{2}{3} + \frac{1}{5} - \frac{1}{2}

What are the solutions to the equation v^2+2v-8=0 ?
The solutions to the equation v^2+2v-8=0 are v=2,v=-4
Find the zeros of v^2+2v-8=0
The zeros of v^2+2v-8=0 are v=2,v=-4