Answer
$\dfrac{2}{3}\pm\dfrac{\sqrt{11}}{3}i$
Work Step by Step
In the form $ax^2+bx+c=0,$ the given equation, $
3t^2-4t=-5
,$ is equivalent to
\begin{align*}
3t^2-4t+5&=0
.\end{align*}
The equation above has $a=
3
$, $b=
-4
$ and $c=
5
$. Using $
x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}
$ or the Quadratic Formula, then
\begin{align*}
t&=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(3)(5)}}{2(3)}
\\\\&=
\dfrac{4\pm\sqrt{16-60}}{6}
\\\\&=
\dfrac{4\pm\sqrt{-44}}{6}
.\end{align*}
Using the properties of radicals, the equation above is equivalent to
\begin{align*}\require{cancel}
t&=\dfrac{4\pm\sqrt{44\cdot(-1)}}{6}
\\\\
t&=\dfrac{4\pm\sqrt{4\cdot11\cdot(-1)}}{6}
\\\\
t&=\dfrac{4\pm\sqrt{4}\cdot\sqrt{11}\cdot\sqrt{-1}}{6}
\\\\
t&=\dfrac{4\pm2\cdot\sqrt{11}\cdot i}{6}
&(\text{use }i=\sqrt{-1})
\\\\
t&=\dfrac{\cancelto24\pm\cancelto12\cdot\sqrt{11}\cdot i}{\cancelto36}
\\\\
t&=\dfrac{2\pm\sqrt{11}\cdot i}{3}
\\\\
t&=\dfrac{2}{3}\pm\dfrac{\sqrt{11}}{3}i
.\end{align*}
Hence, the solution set of $
3t^2-4t=-5
$ is $
\dfrac{2}{3}\pm\dfrac{\sqrt{11}}{3}i
$.
