0=y^2-3y+2

Solution for 0=y^2-3y+2 equation:

0=y^2-3y+2

We move all terms to the left:

0-(y^2-3y+2)=0

We add all the numbers together, and all the variables

-(y^2-3y+2)=0

We get rid of parentheses

-y^2+3y-2=0

We add all the numbers together, and all the variables

-1y^2+3y-2=0

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

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