-2y^2+4y+22=0

Solution for -2y^2+4y+22=0 equation:

-2y^2+4y+22=0

a = -2; b = 4; c = +22;
Δ = b2-4ac
Δ = 42-4·(-2)·22
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8\sqrt{3}}{2*-2}=\frac{-4-8\sqrt{3}}{-4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8\sqrt{3}}{2*-2}=\frac{-4+8\sqrt{3}}{-4}
$