6v^2+v-1=0

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Solution for 6v^2+v-1=0 equation: 



6v^2+v-1=0

a = 6; b = 1; c = -1;
Δ = b2-4ac
Δ = 12-4·6·(-1)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$


$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-5}{2*6}=\frac{-6}{12}
=-1/2
$


$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+5}{2*6}=\frac{4}{12}
=1/3
$

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