The number of non-negative real roots of $2^{x}-x-1$ are 
The number of non-negative real roots of $2^{x}-x-1$ are 

$ a.)\ 0\\
 b.) \ 1 \\
 c.)\ 2 \\
 d.)\ 3 \\$
I don't have any clue.
I have only learned to solve quadratics and cubic equations , 
i haven't been taught to solve such type of equations where
$x$ is written in power.
I look for a short and simple way.
I would prefer a way $\color{red}{\text{without calculus}}$ unless necessary.
I have studied maths up to $12$th grade .Thanks.
 A: Consider the function $$f(x)=2^x-x-1$$ Compute the derivative $$f'(x)=2^x \log (2)-1$$ The derivative cancels for $$x_*=-\frac{\log (\log (2))}{\log (2)}$$ For this value $$f(x_*)=-1+\frac{1}{\log (2)}+\frac{\log (\log (2))}{\log (2)}\approx -0.0860713$$ The second derivative test would show that this is a minimum. So, two real roots.
By inspection, $x=0$ is a root and $x=1$ another. These are the roots.
A: We can be really fancy about this, but I think there's a very simple solution.
$f(x)=0$ is the same as $2^x=x+1$. The left-hand side is an exponential curve; the right-hand side is a line.
Exponential curves can intersect a line at most twice. If you have a good feel for what exponential curves look like, this will be clear, intuitively; if not, you're going to have to use calculus to prove this (i.e., be fancy).
Luckily for us, there are two readily apparent and easy to see solutions: $f(0)=f(1)=0$.
Since we have found two solutions, and we know there can't be any more, we're done--there are exactly two solutions.
A: You could use Bernoulli's inequality, but calculus is used to prove it.
This states

$(1+r)^x \geq 1+rx$ if $x\geq 1$ and $r > -1$ with equality iff $r=0$ or $x=1$.
And $(1+r)^x < 1+rx$ if $0 < x < 1$ and $r > -1$ with equality iff $r=0$.

Rewrite it to $2^x=x+1$.
Insert $r=1$ gives $2^x \geq 1+x$ for $x\geq1$ with equality iff $x=1$.
And $2^x < 1+x$ iff $0 < x < 1$, so no equality here.
We have $x=0$ left, in which case there is equality.
Therefore the only roots are $x=0$ and $x=1$.
A: This is a hit and trial method.
You can graph $f(x)=2^x-x-1$
OR
Since you are looking for non-negative real root(s) and you know that $0$ is the smallest non-negative real number so you first find $f(0)$. You will find that $f(0)=0$. Now with intuition decide what next value of $x$ you should take. Take $x=0.5$, now f(x) is less than zero. $f(1)=0$. Now for any value of $x>1$ you will have $f(x)>0$. Therefore there are no more non-negative real solutions.
Remember : Calculus was made to make calculations easy and short as compared to other alternatives. 
Warning : What I have done is not a proof. Here the problem was easy and there were options so hit and trial may work. If you are taking any exam which have options then this hit and trial may be used. 
If the equation would have been more complicated then I guess that you will love the beauty of calculus because then only you will come to understand why we do calculus.
