2y^2+2y-5=0

Solution for 2y^2+2y-5=0 equation:

2y^2+2y-5=0

a = 2; b = 2; c = -5;
Δ = b2-4ac
Δ = 22-4·2·(-5)
Δ = 44
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:

$\sqrt{\Delta}=\sqrt{44}=\sqrt{4*11}=\sqrt{4}*\sqrt{11}=2\sqrt{11}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{11}}{2*2}=\frac{-2-2\sqrt{11}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{11}}{2*2}=\frac{-2+2\sqrt{11}}{4}
$