What are the solutions to the equation v^2+4v-12=0 ?

The solutions to the equation v^2+4v-12=0 are v=2,v=-6

Find the zeros of v^2+4v-12=0

The zeros of v^2+4v-12=0 are v=2,v=-6

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v^{2} + 4 v - 12 = 0

v = 2, v = - 6

Solution steps

v^{2} + 4 v - 12 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 12 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 12) = 8

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

Separate the solutions

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

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

v = \frac{- 4 - 8}{2 \cdot 1} : - 6

The solutions to the quadratic equation are:

v = 2, v = - 6

h^{2} - 31 h = 0

9 x^{2} - 4 = 0

\frac{x}{2} = (\frac{x}{5})^{2}

x^{2} - 2 x - 1 = 0

v^{2} - 12 v - 81 = - 9

What are the solutions to the equation v^2+4v-12=0 ?
The solutions to the equation v^2+4v-12=0 are v=2,v=-6
Find the zeros of v^2+4v-12=0
The zeros of v^2+4v-12=0 are v=2,v=-6