-3v^2+17v+20=0

Solution for -3v^2+17v+20=0 equation:

-3v^2+17v+20=0

a = -3; b = 17; c = +20;
Δ = b2-4ac
Δ = 172-4·(-3)·20
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-23}{2*-3}=\frac{-40}{-6}
=6+2/3
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+23}{2*-3}=\frac{6}{-6}
=-1
$