Answer
The solution set of this problem, written in standard form, is $$\Big\{\frac{1}{2}\pm\frac{\sqrt6}{2}i\Big\}$$
Work Step by Step
$$4(x^2-x)=-7$$
The equation is not in standard form, so first, we need to bring it back to standard form:
$$4x^2-4x=-7$$
$$4x^2-4x+7=0$$
Now the equation is in standard form, the quadratic formula can be applied.
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
As $a=4, b=-4, c=7$
$$x=\frac{-(-4)\pm\sqrt{(-4)^2-4\times4\times7}}{2\times4}$$
$$x=\frac{4\pm\sqrt{16-112}}{8}$$
$$x=\frac{4\pm\sqrt{-96}}{8}$$
Now we rewrite $\sqrt{-96}=i\sqrt{96}=i\sqrt{16\times6}=4i\sqrt6$
$$x=\frac{4\pm 4i\sqrt6}{8}$$
Finally, we simplify
$$x=\frac{1\pm i\sqrt6}{2}$$
The solution set of this problem, written in standard form, is $$\Big\{\frac{1}{2}\pm\frac{\sqrt6}{2}i\Big\}$$
