12v^2+v-20=0

Solution for 12v^2+v-20=0 equation:

12v^2+v-20=0

a = 12; b = 1; c = -20;
Δ = b2-4ac
Δ = 12-4·12·(-20)
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-31}{2*12}=\frac{-32}{24}
=-1+1/3
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+31}{2*12}=\frac{30}{24}
=1+1/4
$