3y^2+10y-10=0

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

3y^2+10y-10=0

a = 3; b = 10; c = -10;
Δ = b2-4ac
Δ = 102-4·3·(-10)
Δ = 220
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:

$\sqrt{\Delta}=\sqrt{220}=\sqrt{4*55}=\sqrt{4}*\sqrt{55}=2\sqrt{55}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{55}}{2*3}=\frac{-10-2\sqrt{55}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{55}}{2*3}=\frac{-10+2\sqrt{55}}{6}
$