Quadratic Bézier curve points Wikipedia:

A quadratic Bézier curve is the path traced by the function
  $B(t)$, given points $P_0$, $P_1$, and $P_2$.

$$C(t) = \sum _{i=0}^{2}\binom{2}{i} t^i(1-t)^{2-i}P_i$$
$$C(t) = (P_0-2P_1+P_2)t^2+(-2P_0+2P_1)t+P_0 \quad t\in[0,1]$$
What exactly is $P_0$ or $P_1$ or $P_2$ concerning this equation?
Yes they are points. But in my understanding, a point is a pair of numbers (in 2D-space).
Let $P_0$ $=(1,1)$,  $P_1 = (1,7)$ and $P_2=(7,1)$.
What values do you use (and where)?
How do you calculate the Bezier Curve for these points?
 A: Generally, the formula should read (for your special case, the quadratic, $m=2)$    
$C(t) = \sum _{i=0}^m\left(\begin{matrix}m\\i\end{matrix}\right)P_i(1-t)^it^{m-i}$    
In a 2-dimensional environment, with control points $(x_i\mid y_i)\qquad i=\{0,1,2\},$
$m=2.$    
You would need TWO instances of the formula: one for each dimension. The first instance would utilize the $x$
parts of your given control points in place of the Ps. The second instance would utilize the $y$
parts. For each value of $t,$
the two values yielded by these TWO instances would constitute the two coördinates for each point of your desired Bézier curve. Clear, I hope?
For your example we need two coördinates for each of your given points. I have assigned each coördinate a name, as follows:
\begin{array}{c|ccc}
\text{point}&x&y\\
\hline
P_0&P_{0,x}&P_{0,y}\\
P_1&P_{1,x}&P_{1,y}\\
P_2&P_{2,x}&P_{2,y}
\end{array}     
For each value of t, the corresponding point of your desired curve is $(x_t\mid y_t),\quad$ where
$x_t=P_{0,x}(1-t)^2+2P_{1,x}(1-t)t+P_{2,x}t^2$
$y_t=P_{0,y}(1-t)^2+2P_{1,y}(1-t)t+P_{2,y}t^2$    
A: You can use standard operations on points in natural ways.
Adding (or subtracting) two points:  $P_1+P_2$ = $(x_1+x_2,\ y_1+y_2)$
Multiplying (or dividing) a point by a number: $P_1k=(x_1k,\ y_1k)$.
The Bézier curve uses only the standard numerical operators and these two additional abilities.

an example, using your data.
$$C(t) = (P_0-2P_1+P_2)t^2+(-2P_0+2P_1)t+P_0 \quad t\in[0,1]$$
$$C(t) = ((1,1) - 2(1,7) + (7,1))t^2 + (-2(1,1)+2(1,7))t+(1,1)$$
Applying multiplication:
$$C(t) = ((1,1) - (2,14) + (7,1))t^2 + ((-2,-2)+(2,14))t+(1,1)$$
And addition:
$$C(t) = (6,-12)t^2+(0,12)t+(1,1)$$
then we can handle it by components by multiplying and adding some more:
$$C(t) = (6t^2+1, -12t^2+12t+1)$$
Usually you won't see it shown this way, though:  The operations described above are considered already natural and straightforward.
