12v^2+31v+20=0

Solution for 12v^2+31v+20=0 equation:

12v^2+31v+20=0

a = 12; b = 31; c = +20;
Δ = b2-4ac
Δ = 312-4·12·20
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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