Answer
$$x_{1} = \frac{3+ \sqrt {23}i}{4}$$
$$x_{2} = \frac{3- \sqrt {23}i}{4}$$
Work Step by Step
We have $$x(2x-3) = -4$$
First we are going to take the multiplication in order to eliminate the parenthesis $$x*2x + x*(-3) = -4$$ $$2x^2 - 3x = -4$$
As we see we got a quadratic term, a linear term and a constant, so we are going to take all the terms to the left side of the equation and equalize to $0$ $$2x^2 - 3x + 4 = 0$$
Now we got a quadratic equation so we are going to solve it using the quadratic formula. In this case $a = 2$, $b = -3$ and $c = 4$
$$x_{1,2} = \frac{-b ± \sqrt {b^2 - 4ac}}{2a} =\frac{-(-3) ± \sqrt {(-3)^2 - 4(2)(4)}}{2(2)}$$
Doing the operations taking in count the signs law $$x_{1,2} = \frac{3± \sqrt {9 - 32}}{4} =\frac{3 ± \sqrt {-23}}{4}$$
As we know square roots of negative numbers does not exist, so we got and imaginary number $$x_{1,2} = \frac{3± \sqrt {23}i}{4}$$
So for the first value of $x$ we have $$x_{1} = \frac{3+ \sqrt {23}i}{4}$$
And for the second value $$x_{2} = \frac{3- \sqrt {23}i}{4}$$
