3v^2+40v+13=0

Solution for 3v^2+40v+13=0 equation:

3v^2+40v+13=0

a = 3; b = 40; c = +13;
Δ = b2-4ac
Δ = 402-4·3·13
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-38}{2*3}=\frac{-78}{6}
=-13
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+38}{2*3}=\frac{-2}{6}
=-1/3
$