What are the solutions to the equation v^2+2v-14=-12 ?

The solutions to the equation v^2+2v-14=-12 are v=-1+sqrt(3),v=-1-sqrt(3)

Find the zeros of v^2+2v-14=-12

The zeros of v^2+2v-14=-12 are v=-1+sqrt(3),v=-1-sqrt(3)

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v^{2} + 2 v - 14 = - 12

v = - 1 + 3, v = - 1 - 3

Decimal Notation

v = 0.73205 \dots, v = - 2.73205 \dots

Solution steps

v^{2} + 2 v - 14 = - 12

Move

12

to the left side

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 1 \cdot (- 2) = 23

v_{1, 2} = \frac{- 2 \pm 2 3}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- 2 + 2 3}{2 \cdot 1}, v_{2} = \frac{- 2 - 2 3}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

v = - 1 + 3, v = - 1 - 3

∣ 4 y - 1 ∣ \geq 2

50 = - 16 t^{2} + 70 t + 5

x^{2} - 19 x + 42 > - 3 - 5 x

a - 6 = 13

x^{2} + 18 x + 80

What are the solutions to the equation v^2+2v-14=-12 ?
The solutions to the equation v^2+2v-14=-12 are v=-1+sqrt(3),v=-1-sqrt(3)
Find the zeros of v^2+2v-14=-12
The zeros of v^2+2v-14=-12 are v=-1+sqrt(3),v=-1-sqrt(3)