### Instruction: Find all $t$ such that $x-t$ is a factor of $6x^2+13x-5.$

Enter your answer as a list separated by commas.### Response: I know that if $x-t$ is a factor of a polynomial, then the polynomial must be zero when $x=t$.
This means that I can use the quadratic formula to find the values of $x$ that make $6x^2+13x-5=0$ and then set them equal to $t$.
The quadratic formula says that if $ax^2+bx+c=0$, then $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.
In this case, $a=6$, $b=13$, and $c=-5$, so $x=\frac{-13\pm\sqrt{13^2-4(6)(-5)}}{2(6)}$.
Simplifying the expression, I get $x=\frac{-13\pm\sqrt{169+120}}{12}=\frac{-13\pm\sqrt{289}}{12}=\frac{-13\pm 17}{12}$.
Therefore, $x=\frac{4}{12}=\frac{1}{3}$ or $x=\frac{-30}{12}=-\frac{5}{2}$.
These are the values of $x$ that make the polynomial zero, so I can set them equal to $t$ and get $t=\frac{1}{3}$ or $t=-\frac{5}{2}$.