y=y^2-4y+2

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

y=y^2-4y+2

We move all terms to the left:

y-(y^2-4y+2)=0

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1y^2+5y-2=0

a = -1; b = 5; c = -2;
Δ = b2-4ac
Δ = 52-4·(-1)·(-2)
Δ = 17
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{-(5)-\sqrt{17}}{2*-1}=\frac{-5-\sqrt{17}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{17}}{2*-1}=\frac{-5+\sqrt{17}}{-2}
$