Rearrange the function

y(2x^2-x+1) = x^2+x^2+2

(1-2y)x^2+(1+y)x+2-y = 0

For the quadratic equation to have real solutions,

the discriminant must be greater than or equal to 0, that is,

(1+y)^2-4(1-2y)(2-y)\geq 0

Simplifying the inequality gives

Divide the equation by 7 to make the coefficient of leading term 1

On the other hand,

suppose a,b are the minimum and maximum of the function y, that is

a< y < b

Then

(y-a)(y-b)<0

Expand the equation

Comparing coefficients of each terms of (3) with those of (1) results in

a+b = \dfrac{22}{7}

ab = 1

Solving the system of equations for a,b yields

y_{min} = \dfrac{11-6\sqrt{2} }{7}

and

y_{man} = \dfrac{11+6\sqrt{2} }{7}