19y^2+8y-2=0

Solution for 19y^2+8y-2=0 equation:

19y^2+8y-2=0

a = 19; b = 8; c = -2;
Δ = b2-4ac
Δ = 82-4·19·(-2)
Δ = 216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-6\sqrt{6}}{2*19}=\frac{-8-6\sqrt{6}}{38}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+6\sqrt{6}}{2*19}=\frac{-8+6\sqrt{6}}{38}
$