22=6+33y-16y^2

Solution for 22=6+33y-16y^2 equation:

22=6+33y-16y^2

We move all terms to the left:

22-(6+33y-16y^2)=0

We get rid of parentheses

16y^2-33y-6+22=0

We add all the numbers together, and all the variables

16y^2-33y+16=0

a = 16; b = -33; c = +16;
Δ = b2-4ac
Δ = -332-4·16·16
Δ = 65
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-33)-\sqrt{65}}{2*16}=\frac{33-\sqrt{65}}{32}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+\sqrt{65}}{2*16}=\frac{33+\sqrt{65}}{32}
$