First use the fact that the curve passes through the point $(1,2)$. That says that when $x=1$, we have $y=2$. So substitute $x=1$ in the equation $y=ax^2+bx+c$. We get
Similarly, because $(3,7)$ is on the curve, we have
And finally, the third point tells us that
We now have $3$ linear equations in the $3$ unknowns $a$, $b$ and $c$.
From the two equations $9a+3b+c=7$ and $a+b+c=2$, we obtain, by subtraction, that
From the two equations $9a+3b+c=7$ and $a-b+c=1$, we obtain, by subtraction, that $8a+4b=6$. (I deliberately did not subtract the third from the first, that would have made things too easy!)
We have "eliminated" $c$, and we have two equations in the variable $a$ and $b$. Now we will "eliminate" $b$, which again is not the clever thing to do.
So recall we have $8a+2b=5$ and $8a+4b=6$. Multiply both sides of the first equation by $2$. We get $16a+4b=10$. By subtraction, using $8a+4b=6$, we get $8a=4$, and therefore $a=1/2$. Then from $8a+2b=5$ we get $4+2b=5$ and therefore $b=1/2$. Finally, from $a+b+c=2$ we get that $c=1$.