We have: \(x^2-3x+2=x^2-3x+2.25-0.25=x^2-2\times x\times1.5+\left(1.5\right)^2-0.25=\left(x-1.5\right)^2-0.25=0\)

\(\Rightarrow\left(x-1.5\right)^2=0.25\)

\(\Rightarrow x-1.5=\pm0.5\)

\(\Rightarrow x=\left\{1;2\right\}\)

So \(x=\left\{1;2\right\}\)

\(x^2-3x+2=0\)

I consider two cases

Case 1: \(\dfrac{-\left(-3\right)+\sqrt{\left(-3\right)^2-4.1.2}}{2.1}\)

\(=\dfrac{3+\sqrt{\left(-3\right)^2-4.1.2}}{2.1}\)

\(=\dfrac{4}{2.1}\)

= 2

Case 2: \(\dfrac{-\left(-3\right)-\sqrt{\left(-3\right)^2-4.1.2}}{2.1}\)

\(=\dfrac{3-\sqrt{\left(-3\right)^2-4.1.2}}{2.1}\)

\(=\dfrac{2}{2.1}\)

= 1

=> x = {2; 1}

So: x = {2; 1}