An amazing quadratic!

By the Fundamental Theorem of Arithmetic we must have both $|x - 10|$ and $|x - 12|$ as powers of $2$ . Now the only two powers of $2$ that differ by $2$ are $2$ and $4$ . So since we always have that $x - 10 \gt x - 12$ , we can have either

$x - 10 = 4, x - 12 = 2 \Longrightarrow x = 14$ or

$x - 10 = -2, x - 12 = -4 \Longrightarrow x = 8$ .

In both cases $y = 3$ , so there are $\boxed{2}$ ordered pairs, namely $(14,3)$ and $(8,3)$ .

draw the graph of 2^y and x^2-22x+120 they for sure intersect only at two points therefore ordered pair of (x,y) are 2.

A quadratic function intersects 2^y at 2 points knowing that the intersections stay in both the positive x and y directions

This is just like a product of two consecutive even numbers that are both powers of two or the product of their negatives. So we have two sets, the positives and negatives of 2 and 4.

With both 2 and 4 positive, $x=14, y=3$ and with both negative, $x=8, y=3$ .

assume x-10=a and x-12=b when b = a+2 >>>1 a,b should be on the fourmela 2^n >>>>2 from 1 & 2 >>>>> (a,b) = (2,4) , ( -2,-4 ) let x-10 = 4 >>> x=14 >>> y+3 comp >>>