$\text{ The constant }\theta \text{ of Lagrange's theorem for }f\left(x\right)=x^2-2x+3\text{ is }\left[1,\frac{3}{2}\right]\text{ is }$

$\begin{array}{l}\text{ since }f\left(x\right)=x^2-2x+3\\ f^1\left(x\right)=2x-2\end{array}$

$\text{ By LMVT there exists }\theta \in \left(0,\frac{1}{2}\right)\text{ such tha }$

$\begin{array}{r}{f}^1\left(1+\frac{\theta }{2}\right)=\frac{f\left(\frac{3}{2}\right)-f\left(1\right)}{\frac{3}{2}-1}\\ 2\left(1+\frac{\theta }{2}\right)-2=\frac{\frac{9}{4}-2}{\frac{1}{2}}\Rightarrow \theta =\frac{1}{2}=0.5\end{array}$