All real solution of the equation $2^x+3^x+6^x = x^2$ 
Find all real solution of the equation $2^x+3^x+6^x = x^2$

$\bf{My\; Try::}$ Let $$f(x) = 2^x+3^x+6^x-x^2\;,$$ Now Using first Derivative 
$$f'(x) = 2^x\cdot \ln 2+3^x\cdot \ln 3+6^x\cdot \ln 3-2x$$
Now for $x<0\;,$ We get $f'(x)>0,$ So function $f(x)$ is strictly Increasing function.
so it will cut $\bf{X-}$ axis at exactly one point.
So $f(x)=0$ has exactly one root for $x=-1$ for $x<0$
Now How can I calculate no. of real roots for $x\geq 0$
Thanks 
 A: The derivative $f'(x)$ is certainly greater than $h(x)=6^x \ln 6 - 2x.$ [I assume you had a typo when you wrote the term $6^x \ln 3,$ as the derivative of $6^x$ is $6^x \ln 6.$]
Now assume $x \ge 0$ and note $h(0)=\ln 6 >0.$ Also we have
$$h'(x)=6^x (\ln 6)^2 -2 > (3.2)\cdot 6^x -2\ge 1.2 >0,$$
using the underestimate $3.2$ for $(\ln 6)^2$ and the fact that for nonnegative $x$ one has $6^x \ge 1.$
Thus we have both $h(0)>0$ and $h$ increasing on $[0,\infty),$ and can conclude as desired that $h(x) \ge 0$ for $x \ge 0.$ Combining this with your result for negative $x$ one gets $f$ increasing on $\mathbb{R}$ so the only zero of $f$ is at $x=-1.$
A: I'm using the f(x) you defined.
You can numerically verify that f(0) and f'(0) are both >0.
You can analytically verify that all higher derivatives of f(x) will be >0 at x=0.
So if f(x) and all of it's derivatives are >0 at x=0, it is not possible for f(x) to dip down to 0 for any x>0
A: Your derivative is equal to 
$f'(x) = \ln{ 2^{2^x}+ \ln 3^{3^x}+ \ln 6^{6^x}-\ln e^{-2x}=\ln2^{2^x}3^{3^x}6^{6^x}e^{2x}}$ 
Without calculating $f '(0)$, it is clear that it is  positive and increasing so positive for $x\ge 0$. Hence $f$ is  increasing for $x\gt 0$; besides $f(0)=3$ thus none positive solution.
NOTE.-A comment induced me to edit what was right for something wrong. Again I put what I wrote first.
