2v^2+40v+38=0

Solution for 2v^2+40v+38=0 equation:

2v^2+40v+38=0

a = 2; b = 40; c = +38;
Δ = b2-4ac
Δ = 402-4·2·38
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-36}{2*2}=\frac{-76}{4}
=-19
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+36}{2*2}=\frac{-4}{4}
=-1
$