Answer
$\left\{1\pm\dfrac{\sqrt{3}}{3}i\right\}$
Work Step by Step
In the form $ax^2+bx+c=0,$ the given equation, $
3t^2-6t=-4
$, is equivalent to
\begin{align*}
3t^2-6t+4&=0
.\end{align*}
The equation above has $a=
3
$, $b=
-6$ and $c=
4
$. Using $
x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}
$ or the Quadratic Formula, then
\begin{align*}
t&=\dfrac{-(-6)\pm\sqrt{(-6)^2-4(3)(4)}}{2(3)}
\\\\&=
\dfrac{6\pm\sqrt{36-48}}{6}
\\\\&=
\dfrac{6\pm\sqrt{-12}}{6}
.\end{align*}
Using the properties of radicals, the equation above is equivalent to
\begin{align*}\require{cancel}
t&=\dfrac{6\pm\sqrt{12\cdot(-1)}}{6}
\\\\
t&=\dfrac{6\pm\sqrt{12}\cdot\sqrt{-1}}{6}
\\\\
t&=\dfrac{6\pm\sqrt{4\cdot3}\cdot\sqrt{-1}}{6}
\\\\
t&=\dfrac{6\pm\sqrt{4}\cdot\sqrt{3}\cdot\sqrt{-1}}{6}
\\\\
t&=\dfrac{6\pm2\cdot\sqrt{3}\cdot i}{6}
&(\text{use }i=\sqrt{-1})
\\\\
t&=\dfrac{\cancelto36\pm\cancelto12\cdot\sqrt{3}\cdot i}{\cancelto36}
\\\\
t&=\dfrac{3\pm\sqrt{3}\cdot i}{3}
\\\\
t&=\dfrac{3}{3}\pm\dfrac{\sqrt{3}\cdot i}{3}
\\\\
t&=1\pm\dfrac{\sqrt{3}}{3}i
.\end{align*}
Hence, the solution set of $
3t^2-6t=-4
$ is $
\left\{1\pm\dfrac{\sqrt{3}}{3}i\right\}
$.
