Let $f(x) = 2x^2 + nx - 6$ and $g(x) = mx^2 + 2x - 4$? The functions are combined to form the new functions $h(x) = f(x) - g(x)$. Points $(2, 4)$ and $(-3, 17)$ are given to satisfy the new function. Determine the values of $m$ and $n$.
I know these formulas but I don't know what is $m$ and $n$ values are.
$$h(x)=(2-m)(x^2)+(n-2)(x)-2$$
$$h(2)=4=(2-m)(2^2)+(n-2)(2)-2$$
$$h(-3)=17=(2-m)(-3^2)+(n-2)(-3)-2$$
Help, please.

Just continue... But be careful of arithmetic
$h(x) = f(x) - g(x) = (2-m)x^2 + (n-2)x -2$
$h(2) = 4(2-m) + 2(n-2)-2 = -4m + 2n +2 = 4$.
so simplify  $-4m + 2n = 2$ and $-2m + n = 1$
$h(-3) = 9(2-m)-3(n-2) -2 = -9m -3n+22 = 17$
so simplify $-9m -3n = -5$ so $3m +n = \frac 53$
Now just solve for $n$ and $m$.
Simple substitution $n = 2m + 1$ and so $3m + 2m+1 = \frac 53$ so
$5m = \frac 23$ and $m = \frac 2{15}$
ANd $n = 2\cdot\frac 2{15} + 1 = \frac {19}{15}$
Let's verify as I've seen answers all over the map.
$f(x) = 2x^2 +\frac {19}{15}x -6$ so $f(2)= 8+\frac {38}{15}-6 = 4\frac 8{15}$. and $f(-3) = 18-\frac {19}5-6=12-3\frac 45= 8\frac 15$.
Ahd $h(x) = \frac 2{15}x^2 +2x -4$ so $g(2) = \frac 8{15}+4+4=\frac 8{15}$ and $g(-3)=\frac {18}{15}-6-4= -10+1\frac 15= -8\frac 45$.
And $h(2)=4\frac 8{15}-\frac 8{15} = 4$. And $h(-3) = 8\frac 15 - (-8\frac 45)=17$.
Hmmm.... what do you know.  My arithmetic was correct and everyone else's was wrong.
That's got to be a first.
