3y^2+28y-130=0

Solution for 3y^2+28y-130=0 equation:

3y^2+28y-130=0

a = 3; b = 28; c = -130;
Δ = b2-4ac
Δ = 282-4·3·(-130)
Δ = 2344
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:

$\sqrt{\Delta}=\sqrt{2344}=\sqrt{4*586}=\sqrt{4}*\sqrt{586}=2\sqrt{586}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-2\sqrt{586}}{2*3}=\frac{-28-2\sqrt{586}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+2\sqrt{586}}{2*3}=\frac{-28+2\sqrt{586}}{6}
$