3y^2+17y-3=0

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

3y^2+17y-3=0

a = 3; b = 17; c = -3;
Δ = b2-4ac
Δ = 172-4·3·(-3)
Δ = 325
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:

$\sqrt{\Delta}=\sqrt{325}=\sqrt{25*13}=\sqrt{25}*\sqrt{13}=5\sqrt{13}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-5\sqrt{13}}{2*3}=\frac{-17-5\sqrt{13}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+5\sqrt{13}}{2*3}=\frac{-17+5\sqrt{13}}{6}
$