What are the solutions to the equation v^2+v-6=0 ?

The solutions to the equation v^2+v-6=0 are v=2,v=-3

Find the zeros of v^2+v-6=0

The zeros of v^2+v-6=0 are v=2,v=-3

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v^{2} + v - 6 = 0

v = 2, v = - 3

Solution steps

v^{2} + v - 6 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

v = 2, v = - 3

3 x^{2} - 5 x + 2 = 4 - 6 x

(4 y + 10) (y - 3 y) = 5

x^{2} + 8 x - 4 = 0

3 x^{2} - 4 = - x^{2}

2 - x^{2} + x = 0

What are the solutions to the equation v^2+v-6=0 ?
The solutions to the equation v^2+v-6=0 are v=2,v=-3
Find the zeros of v^2+v-6=0
The zeros of v^2+v-6=0 are v=2,v=-3