2y^2+2y=22

Solution for 2y^2+2y=22 equation:

2y^2+2y=22

We move all terms to the left:

2y^2+2y-(22)=0

a = 2; b = 2; c = -22;
Δ = b2-4ac
Δ = 22-4·2·(-22)
Δ = 180
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:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6\sqrt{5}}{2*2}=\frac{-2-6\sqrt{5}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6\sqrt{5}}{2*2}=\frac{-2+6\sqrt{5}}{4}
$