10v^2+19v+6=0

Solution for 10v^2+19v+6=0 equation:

10v^2+19v+6=0

a = 10; b = 19; c = +6;
Δ = b2-4ac
Δ = 192-4·10·6
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-11}{2*10}=\frac{-30}{20}
=-1+1/2
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+11}{2*10}=\frac{-8}{20}
=-2/5
$