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: **To OP: The full rquation is not a quadratic equation, it is a transcendental one.^^
Let $f(x)=x^2+x-3-\ln(x+2) \implies f(-2)\rightarrow  \infty>0,  f(-1)=-3 <0, f(0)=-3=\ln 2<0$ so one real root in $(-2,-1)$
Next, $f(2)=4+2-3-\ln 4=3-1.38>0, f(1)=-1-\ln 3<0$ so one real root in $(1,2)$. So at least two real roots by internediate theorem.
Further, write it  as $y=x^2+x-3=\ln(x+2)=y$ first one is is prarabola and the other one is shifted logarithmic which keeps increasing and leaves out the parabola not to cut it agai. So only two real roots. See the logarithmis and parabolix cyrves in the Fig. below.

A: Hint: as you correctly identified, you can reduce this to finding a zero of a function. What value does this function have at zero and what is its derivative?
A: I have an even simpler solution using Bolzano's theorem. We know that $\ln 2\approx 0.693$ so take $f(x)=x^{2}+x-3-\ln(x+2)$.
Now observe $f(0)=-3.693$ (approx) $<0$ and $f(8)=69-\ln(10)>0$, so by Bolzao's theorem we have a root in $[0,8]\subset [0,\infty)$.
A: 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.
A: You have misused the quadratic formula by considering $3+\ln (x+2)$ as a constant.
You may use Rolle's Theorem to solve your problem. 
Let $$y=x^2+x-3-\ln(x+2)$$ and show that $y'$ does not have a positive zero. 
That will show that you can not have more than one positive zero for the function $$y=x^2+x-3-\ln(x+2)$$
Then you can show that the equation does have one positive zero using the intermediate value theorem.  
A: This is not a  quadratic equation since there is  $\ln(x+2)$.
Anyway, you're not asked to find the solution, but show there's exactly one.
Hint: consider the function 
$$f(x)=x^2+x-\bigl(2+\ln(x+2)\bigr),$$
observe that it is increasing on $[0,+\infty)$ and apply the (extended) intermediate value theorem.
A: 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 $f(2)>0$
