$(2x+5)^{2}$ =$(x+2)^{2}$
⇔$(2x+5)^{2}$-$(x+2)^{2}$ =0
⇔4$x^{2}$ +20+25-$x^{2}$ -4x-4=0
⇔3$x^{2}$ +16x+21=0
⇔(x+3)(3x+ 7)=0
⇔\(\left[ \begin{array}{l}x+3=0\\3x+7=0\end{array} \right.\)=0
⇔\(\left[ \begin{array}{l}x=-3\\x=\frac{-7}{3} \end{array} \right.\) .
Vậy x∈(-3;$\frac{-7}{3}$ )