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

The solutions to the equation 2v^2+2v-6=0 are v=(-1+sqrt(13))/2 ,v=-(1+sqrt(13))/2

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

The zeros of 2v^2+2v-6=0 are v=(-1+sqrt(13))/2 ,v=-(1+sqrt(13))/2

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2 v^{2} + 2 v - 6 = 0

v = \frac{- 1 + 13}{2}, v = - \frac{1 + 13}{2}

Decimal Notation

v = 1.30277 \dots, v = - 2.30277 \dots

Solution steps

2 v^{2} + 2 v - 6 = 0

Solve with the quadratic formula

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

2^{2} - 4 \cdot 2 (- 6) = 213

v_{1, 2} = \frac{- 2 \pm 2 13}{2 \cdot 2}

Separate the solutions

v_{1} = \frac{- 2 + 2 13}{2 \cdot 2}, v_{2} = \frac{- 2 - 2 13}{2 \cdot 2}

v = \frac{- 2 + 2 13}{2 \cdot 2} : \frac{- 1 + 13}{2}

v = \frac{- 2 - 2 13}{2 \cdot 2} : - \frac{1 + 13}{2}

The solutions to the quadratic equation are:

v = \frac{- 1 + 13}{2}, v = - \frac{1 + 13}{2}

72 a^{5} b^{3} c - 32 a^{3} b^{2} c^{2}

x^{2} + 10 x + 16

(\frac{( x ^{6} )}{3})^{3}

∣ x + 1 ∣ < 2

a^{5} + b^{5}

What are the solutions to the equation 2v^2+2v-6=0 ?
The solutions to the equation 2v^2+2v-6=0 are v=(-1+sqrt(13))/2 ,v=-(1+sqrt(13))/2
Find the zeros of 2v^2+2v-6=0
The zeros of 2v^2+2v-6=0 are v=(-1+sqrt(13))/2 ,v=-(1+sqrt(13))/2