Question 382·Hard·Math

The whole input to $f$ is $t^2-7t$. By the zero-product property, $(x-2)(x+1)(x-8)$ equals $0$ when one factor equals $0$. So the equation splits into

Type $f(x)=(x-2)(x+1)(x-8)$, then $f(x^2-7x)=0$. Desmos uses $x$ on its horizontal axis, so its $x$-values stand for $t$. It draws six vertical lines; click them to read values near $-1$, $-0.275$, $0.146$, $6.854$, $7.275$, and $8$. Each of the three quadratic equations can have at most two real solutions, so each has two.

Let $k$ stand for one of the possible inputs: $2$, $-1$, or $8$. Subtract $k$ to put each quadratic, an equation with $t^2$, in the same form: $t^2-7t-k=0$. Only the constant term changes. The coefficients of $t^2$ and $t$ are always $1$ and $-7$.

The sum-of-roots rule says the two solutions of $at^2+bt+c=0$ add to $-b/a$. Here $a=1$ and $b=-7$, regardless of $k$. Substitute those coefficients: $r+s=-\frac{-7}{1}$. Simplify: $r+s=7$. Each possible input of $f$ produces a pair of $t$-values with the same sum. Use this exact sum, not the rounded graph labels.

The three possible inputs are different, so their pairs can't share a $t$-value. There are three pairs among the six real values of $t$, giving a total of $3(7)=21$. The sum of all possible values of $t$ is $21$. Grid in 21.