3y^2=22y-32

Solution for 3y^2=22y-32 equation:

3y^2=22y-32

We move all terms to the left:

3y^2-(22y-32)=0

We get rid of parentheses

3y^2-22y+32=0

a = 3; b = -22; c = +32;
Δ = b2-4ac
Δ = -222-4·3·32
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-10}{2*3}=\frac{12}{6}
=2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+10}{2*3}=\frac{32}{6}
=5+1/3
$