2y^2-18y+22=0

Solution for 2y^2-18y+22=0 equation:

2y^2-18y+22=0

a = 2; b = -18; c = +22;
Δ = b2-4ac
Δ = -182-4·2·22
Δ = 148
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:

$\sqrt{\Delta}=\sqrt{148}=\sqrt{4*37}=\sqrt{4}*\sqrt{37}=2\sqrt{37}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{37}}{2*2}=\frac{18-2\sqrt{37}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{37}}{2*2}=\frac{18+2\sqrt{37}}{4}
$