-3y^2+2y+17=0

Solution for -3y^2+2y+17=0 equation:

-3y^2+2y+17=0

a = -3; b = 2; c = +17;
Δ = b2-4ac
Δ = 22-4·(-3)·17
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{13}}{2*-3}=\frac{-2-4\sqrt{13}}{-6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{13}}{2*-3}=\frac{-2+4\sqrt{13}}{-6}
$