2y^2+2y-6=0

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

2y^2+2y-6=0

a = 2; b = 2; c = -6;
Δ = b2-4ac
Δ = 22-4·2·(-6)
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{13}}{2*2}=\frac{-2-2\sqrt{13}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{13}}{2*2}=\frac{-2+2\sqrt{13}}{4}
$