### 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: To find the values of $t$ that make $x-t$ a factor of $6x^2+13x-5$, I need to use the factor theorem.
This theorem says that if $f(x)$ is a polynomial, then $x-a$ is a factor of $f(x)$ if and only if $f(a)=0$.
So, in this case, $x-t$ is a factor of $6x^2+13x-5$ if and only if $6x^2+13x-5=0$ when $x=t$.
This means I need to solve the equation $6t^2+13t-5=0$ for $t$.
This is a quadratic equation, so I can use the quadratic formula to find its roots.
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 plugging these values into the formula, I get $t=\frac{-13\pm\sqrt{13^2-4(6)(-5)}}{2(6)}$.
Simplifying the expression, I get $t=\frac{-13\pm\sqrt{169+120}}{12}$.
Further simplifying, I get $t=\frac{-13\pm\sqrt{289}}{12}$.
Since $\sqrt{289}=17$, I can write $t=\frac{-13\pm 17}{12}$.
This gives me two possible values for $t$: $t=\frac{-13+17}{12}=\frac{4}{12}=\frac{1}{3}$ or $t=\frac{-13-17}{12}=\frac{-30}{12}=\frac{-5}{2}$.
Therefore, the values of $t$ that make $x-t$ a factor of $6x^2+13x-5$ are $\frac{1}{3}$ and $\frac{-5}{2}$.