Prove that the equation:$x^{2}+x=3+\ln(x+2) $ has only one real solution at $[0,\infty ) $ My attempt:
$x^{2}+x-(3+\ln(x+2))=0$
$x_{1,2}=\dfrac{-1\pm \sqrt{1+12+4\ln(x+2)}}{2} \Rightarrow 13+4\ln(x+2)>0 $
$ \Rightarrow 4\ln(x+2)>-13 $
$\ln(x+2)>-\frac{13}{4} $
$x+2>e^{-\frac{13}{4}} $
$x>e^{-\frac{13}{4}}+2 $
What did I do wrong?
Edit: Thank you all for the help:)

A simple proof is that $x^2+x$ and $3+\ln (x+2)$ are convex and concave functions respectively on $[0,\infty)$ and both strictly increasing. Also$$x^2+x|_{x=10}<3+\ln(x+2)|_{x=0}$$and $$x^2+x|_{x=0}>3+\ln(x+2)|_{x=10}$$hence aa unique root exists within $(0,10)$ and the proof is complete.
