Answer
$x=\dfrac{1-i\sqrt{35}}{6}
\text{ OR }
x=\dfrac{1+i\sqrt{35}}{6}
$
Work Step by Step
In the given equation,
\begin{align*}
-3x^2+x-3=0
,\end{align*} $a=
-3
,$ $b=
1
,$ and $c=
-3
.$ Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then
\begin{align*}\require{cancel}x&=
\dfrac{-1\pm\sqrt{1^2-4(-3)(-3)}}{2(-3)}
\\\\&=
\dfrac{-1\pm\sqrt{1-36}}{-6}
\\\\&=
\dfrac{-1\pm\sqrt{-35}}{-6}
\\\\&=
\dfrac{-1\pm\sqrt{-1}\cdot\sqrt{35}}{-6}
\\\\&=
\dfrac{-1\pm i\sqrt{35}}{-6}
&\text{ (use $i=\sqrt{-1}$)}
\\\\&=
\dfrac{\cancel{-1}^1\pm i\sqrt{35}}{\cancel{-6}^6}
&\text{ (divide by $-1$)}
\\\\&=
\dfrac{1\pm i\sqrt{35}}{6}
.\end{align*}
The solutions are $
x=\dfrac{1-i\sqrt{35}}{6}
\text{ and }
x=\dfrac{1+i\sqrt{35}}{6}
.$
