19y^2+6y-39=0

Solution for 19y^2+6y-39=0 equation:

19y^2+6y-39=0

a = 19; b = 6; c = -39;
Δ = b2-4ac
Δ = 62-4·19·(-39)
Δ = 3000
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:

$\sqrt{\Delta}=\sqrt{3000}=\sqrt{100*30}=\sqrt{100}*\sqrt{30}=10\sqrt{30}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-10\sqrt{30}}{2*19}=\frac{-6-10\sqrt{30}}{38}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+10\sqrt{30}}{2*19}=\frac{-6+10\sqrt{30}}{38}
$