What are the solutions to the equation 2v^2+4v-7=0 ?

The solutions to the equation 2v^2+4v-7=0 are v=(-2+3sqrt(2))/2 ,v=-(sqrt(2)+3)/(sqrt(2))

Find the zeros of 2v^2+4v-7=0

The zeros of 2v^2+4v-7=0 are v=(-2+3sqrt(2))/2 ,v=-(sqrt(2)+3)/(sqrt(2))

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2 v^{2} + 4 v - 7 = 0

v = \frac{- 2 + 3 2}{2}, v = - \frac{2 + 3}{2}

Decimal Notation

v = 1.12132 \dots, v = - 3.12132 \dots

Solution steps

2 v^{2} + 4 v - 7 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 2 ( - 7 )}{2 \cdot 2}

4^{2} - 4 \cdot 2 (- 7) = 62

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

Separate the solutions

v_{1} = \frac{- 4 + 6 2}{2 \cdot 2}, v_{2} = \frac{- 4 - 6 2}{2 \cdot 2}

v = \frac{- 4 + 6 2}{2 \cdot 2} : \frac{- 2 + 3 2}{2}

v = \frac{- 4 - 6 2}{2 \cdot 2} : - \frac{2 + 3}{2}

The solutions to the quadratic equation are:

v = \frac{- 2 + 3 2}{2}, v = - \frac{2 + 3}{2}

4 - 2 (x + 5) = - 3 (x - 1)

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

x^{8} - x^{4}

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

- 2 x - 4 = x + 2

What are the solutions to the equation 2v^2+4v-7=0 ?
The solutions to the equation 2v^2+4v-7=0 are v=(-2+3sqrt(2))/2 ,v=-(sqrt(2)+3)/(sqrt(2))
Find the zeros of 2v^2+4v-7=0
The zeros of 2v^2+4v-7=0 are v=(-2+3sqrt(2))/2 ,v=-(sqrt(2)+3)/(sqrt(2))