<s>[INST] 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:)
 [/INST] Consider that you look for the zero of function
$$f(x)=x^2+x-\log (x+2)-3$$ Its derivatives are
$$f'(x)=2 x+1-\frac{1}{x+2} \qquad \text{and} \qquad f''(x)=\frac{1}{(x+2)^2}+2 \quad >0 \forall x$$
The first derivative cancels at negative values. So the function is invreasing and only one root.
By inspection, $f(1)<0$ and $f(2)>0$
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