The value of $\theta$  in Lagrange’s mean value theorem for $f\left(x\right)=l{x}^2+mx+n\left(l\ne 0\right)$  on  $\left[a,b\right]$  is____

$$\begin{gathered} \boxed{f^{\text{'}}\left(a+\theta h\right)=\frac{f\left(a+h\right)-f\left(a\right)}{h}}\boxed{ } \\ \Rightarrow 2l\left(a+\theta h\right)+m=\frac{\left(l{\left(a+h\right)}^2+m\left(a+h\right)+n\right)-\left(l{a}^2+ma+n\right)}{h} \\ \therefore 2l\left(a+\theta h\right)+m=\frac{l{h}^2+mh+2lah}{h} \\ \Rightarrow 2la+2l\theta h+m=lh+m+2la \\ \Rightarrow 2l\theta h=lh \\ \Rightarrow \theta =\frac{1}{2} \end{gathered}$$