An algebra problem by Mohammad Oladzad

Here we can use a bit calculus Given $y=2x^2-2x+2$ ,in order to get minimum value of $y$ we have to show that $\frac{d^2y}{d x^2}>0$

Now let us see :-

$\frac{dy}{dx}=4x-2$ and $\frac{d^2y}{d x^2}=4 >0$ so $y$ will be minimum when $\frac{dy}{dx}=4x-2=0$ ie, $x=\frac{1}{2}$ put $x=\frac{1}{2}$ in the above equation than $y=2×\frac{1}{2^2}-2×\frac{1}{2}+2$ which is $\boxed{1.50}$

$2x^2-2x+2 = 2(x^2-x+1) = 2(x^2-2(\frac{1}{2})x+(\frac{1}{2})^2+\frac{3}{4}) = 2(x+\frac{1}{2})^2+\frac{3}{2}$

Provided that for all real number $n$ , $n \geq 0$ , then $2(x+\frac{1}{2})^2 \geq 0$ $2(x+\frac{1}{2})^2+\frac{3}{2} \geq \frac{3}{2}$ $2x^2-2x+2 \geq \frac{3}{2}$