15v^2+v-6=0

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Solution for 15v^2+v-6=0 equation:



15v^2+v-6=0
a = 15; b = 1; c = -6;
Δ = b2-4ac
Δ = 12-4·15·(-6)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-19}{2*15}=\frac{-20}{30} =-2/3 $
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+19}{2*15}=\frac{18}{30} =3/5 $

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