Answer
$t=\left\{ -2-i\sqrt{7},-2+i\sqrt{7} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the solutions of the given equation, $
t^2+4t+11=0
,$ use the Quadratic Formula.
$\bf{\text{Solution Details:}}$
The quadratic equation above has $a=
1
, b=
4
, c=
11
.$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then
\begin{array}{l}\require{cancel}
t=\dfrac{-4\pm\sqrt{4^2-4(1)(11)}}{2(1)}
\\\\
t=\dfrac{-4\pm\sqrt{16-44}}{2}
\\\\
t=\dfrac{-4\pm\sqrt{-28}}{2}
.\end{array}
Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
t=\dfrac{-4\pm\sqrt{-1}\cdot\sqrt{28}}{2}
.\end{array}
Since $i=\sqrt{-1},$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
t=\dfrac{-4\pm i\sqrt{28}}{2}
.\end{array}
Simplifying the radicand by writing it as an expression that contains a factor that is a perfect square of the index and then extracting the root of that factor, the equation above is equivalent to
\begin{array}{l}\require{cancel}
t=\dfrac{-4\pm i\sqrt{4\cdot7}}{2}
\\\\
t=\dfrac{-4\pm i\sqrt{(2)^2\cdot7}}{2}
\\\\
t=\dfrac{-4\pm i(2)\sqrt{7}}{2}
\\\\
t=\dfrac{-4\pm 2i\sqrt{7}}{2}
.\end{array}
Cancelling the common factor in each term results to
\begin{array}{l}\require{cancel}
t=\dfrac{\cancel2(-2)\pm \cancel2(1)i\sqrt{7}}{\cancel2(1)}
\\\\
t=-2\pm i\sqrt{7}
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
t=-2- i\sqrt{7}
\\\\\text{OR}\\\\
t=-2+i\sqrt{7}
.\end{array}
Hence, $
t=\left\{ -2-i\sqrt{7},-2+i\sqrt{7} \right\}
.$
