#### Answer

$(2t+11u)(3t-7u)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $
6t^2+19tu-77u^2
,$ use the factoring of trinomials in the form $ax^2+bx+c.$
$\bf{\text{Solution Details:}}$
In the trinomial expression above, $a=
6
,b=
19
,\text{ and } c=
-77
.$ Using the factoring of trinomials in the form $ax^2+bx+c,$ the two numbers whose product is $ac=
6(-77)=-462
$ and whose sum is $b$ are $\left\{
33,-14
\right\}.$ Using these two numbers to decompose the middle term results to
\begin{array}{l}\require{cancel}
6t^2+33tu-14tu-77u^2
.\end{array}
Using factoring by grouping, the expression above is equivalent to
\begin{array}{l}\require{cancel}
(6t^2+33tu)-(14tu+77u^2)
\\\\=
3t(2t+11u)-7u(2t+11u)
\\\\=
(2t+11u)(3t-7u)
.\end{array}