3y^2+6y-10=0

Solution for 3y^2+6y-10=0 equation:

3y^2+6y-10=0

a = 3; b = 6; c = -10;
Δ = b2-4ac
Δ = 62-4·3·(-10)
Δ = 156
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:

$\sqrt{\Delta}=\sqrt{156}=\sqrt{4*39}=\sqrt{4}*\sqrt{39}=2\sqrt{39}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{39}}{2*3}=\frac{-6-2\sqrt{39}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{39}}{2*3}=\frac{-6+2\sqrt{39}}{6}
$