Answer
$x=\left\{ \dfrac{7-3i\sqrt{3}}{4},\dfrac{7+ 3i\sqrt{3}}{4} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the non real complex solutions of the given equation, $
(4m-7)^2=-27
,$ take the square root of both sides (Square Root Property). Then use the properties of radicals and use $i=\sqrt{-1}.$ Finally, simplify the radical and isolate the variable.
$\bf{\text{Solution Details:}}$
Taking the square root of both sides (Square Root Property), the equation above is equivalent to
\begin{array}{l}\require{cancel}
4m-7=\pm\sqrt{-27}
.\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}
4m-7=\pm\sqrt{-1}\cdot\sqrt{27}
.\end{array}
Using $i=\sqrt{-1},$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
4m-7=\pm i\sqrt{27}
.\end{array}
Writing the radicand as an expression that contains a factor that is a perfect power of the given index and then extracting the root of that factor, the equation above is equivalent to
\begin{array}{l}\require{cancel}
4m-7=\pm i\sqrt{9\cdot3}
\\\\
4m-7=\pm i\sqrt{(3)^2\cdot3}
\\\\
4m-7=\pm i(3)\sqrt{3}
\\\\
4m-7=\pm 3i\sqrt{3}
.\end{array}
Using the properties of equality to isolate the variable, the equation above is equivalent to
\begin{array}{l}\require{cancel}
4m=7\pm 3i\sqrt{3}
\\\\
m=\dfrac{7\pm 3i\sqrt{3}}{4}
.\end{array}
Hence, $
x=\left\{ \dfrac{7-3i\sqrt{3}}{4},\dfrac{7+ 3i\sqrt{3}}{4} \right\}
.$