Find the roots

According to the question : $3^x=x^2-x+2$

The R.H.S. of the equation i.e. $x^2-x+2$ has minimum value is $\frac{7}{4}$ which is at $x=1/2$

At $x=1/2$ the L.H.S. of the equation $3^x$ has value $1.732$ which is lower than $7/4$ .So, at $x=1/2$ $3^x$ is lower than $x^2-x+2$ .

Taking the difference of their rate of increasing (i.e. derivative) $3^x ln3-(2 x-1)$

It is easily seen that ${3^x ln3-(2 x-1)} >0$

At $x=1/2$ $3^x$ has alower value than $x^2-x+2$ .But rate of increasing show that $3^x$ increase faster than $x^-x+2$ .

So, they must intersect at exactly one point and then increases.

So, there exists one solution.

Make the graphs of $y=3^{x}$ and $y=x^2-x+2$ and check for the number of intersections.

@Chew-Seong Cheong please tell me the answer of this question. and the solution too.