3y2+19y+6=0

Solution for 3y2+19y+6=0 equation:

3y^2+19y+6=0

a = 3; b = 19; c = +6;
Δ = b2-4ac
Δ = 192-4·3·6
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-17}{2*3}=\frac{-36}{6}
=-6
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+17}{2*3}=\frac{-2}{6}
=-1/3
$