Answer
The solution set for the quadratic equation, $x\left( 2x-3 \right)=-4$ is $\left\{ \frac{3}{4}\pm \frac{\sqrt{23}}{4}i \right\}$.
Work Step by Step
Consider quadratic equation,
$x\left( 2x-3 \right)=-4$
Convert it to the standard form. Add 4 to both sides and multiply x with terms in bracket.
$\begin{align}
& x\left( 2x-3 \right)+4=-4+4 \\
& 2{{x}^{2}}-3x+4=0
\end{align}$
Compare the equation with the standard quadratic equation $a{{x}^{2}}+bx+c=0\text{ , }\left( a\ne 0 \right)$.
Here, $a=2,\text{ }b=-3\text{ and }c=4$
Apply quadratic formula
$x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Substitute 2 for a, $-3$ for b and 4 for c.
$x=\frac{-\left( -3 \right)\pm \sqrt{{{\left( -3 \right)}^{2}}-4\left( 2 \right)\left( 4 \right)}}{2\left( 2 \right)}$
Simplify the radical.
$\begin{align}
& x=\frac{-\left( -3 \right)\pm \sqrt{{{\left( -3 \right)}^{2}}-4\left( 2 \right)\left( 4 \right)}}{2\left( 2 \right)} \\
& =\frac{3\pm \sqrt{9-32}}{4} \\
& =\frac{3\pm \sqrt{-23}}{4}
\end{align}$
Rewrite $\frac{3\pm \sqrt{-23}}{4}$ as $\frac{3\pm \sqrt{23}\sqrt{-1}}{4}$
As $i=\sqrt{-1}$
Therefore,
$\begin{align}
& x=\frac{3\pm i\sqrt{23}}{4} \\
& =\frac{3}{4}\pm \frac{\sqrt{23}}{4}i
\end{align}$
The solutions of the quadratic are complex conjugates of each other.
Hence, the solution set for the quadratic equation $x\left( 2x-3 \right)=-4$ is $\left\{ \frac{3}{4}\pm \frac{\sqrt{23}}{4}i \right\}$.
