Answer
$\left\{ \dfrac{3- i\sqrt{87}}{8},\dfrac{3+ i\sqrt{87}}{8} \right\}$
Work Step by Step
The standard form of the given equation, $
6=-4x^2+3x
,$ is
\begin{array}{l}\require{cancel}
4x^2-3x+6=0
.\end{array}
Using $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), the solutions of the quadratic equation, $
4x^2-3x+6=0
,$ are
\begin{array}{l}\require{cancel}
\dfrac{-(-3)\pm\sqrt{(-3)^2-4(4)(6)}}{2(4)}
\\\\=
\dfrac{3\pm\sqrt{9-96}}{8}
\\\\=
\dfrac{3\pm\sqrt{-87}}{8}
\\\\=
\dfrac{3\pm\sqrt{-1}\cdot\sqrt{87}}{8}
\\\\=
\dfrac{3\pm i\sqrt{87}}{8}
.\end{array}
Hence, the solutions are $
\left\{ \dfrac{3- i\sqrt{87}}{8},\dfrac{3+ i\sqrt{87}}{8} \right\}
.$
