Answer
$\left\{\dfrac{1\pm\sqrt{41}}{2}\right\}$
Work Step by Step
In the form $ax^2+bx+c=0,$ the given equation, $
(t+3)(t-4)=-2
,$ is equivalent to
\begin{align*}
t(t)+t(-4)+3(t)+3(-4)&=-2
&(\text{use }(a+b)(c+d)=ac+ad+bc+bd)
\\
t^2-4t+3t-12&=-2
\\
t^2+(-4t+3t)+(-12+2)&=0
\\
t^2-t-10&=0
.\end{align*}
The equation above has
\begin{align*}a=
1
,b=
-1
,\text{ and }c=
-10
.\end{align*}
Using $
x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}
$ or the Quadratic Formula, then
\begin{align*}
x&=\dfrac{-(-1)\pm\sqrt{(-1)^2-4(1)(-10)}}{2(1)}
\\\\&=
\dfrac{1\pm\sqrt{1+40}}{2}
\\\\&=
\dfrac{1\pm\sqrt{41}}{2}
.\end{align*}
Hence, the solution set of $
(t+3)(t-4)=-2
$ is $
\left\{\dfrac{1\pm\sqrt{41}}{2}\right\}
$.
