Quadratic equation

The quadratic equation is given by:

$\begin{aligned} \left(x - \frac {2+3i}2\right) \left(x - \frac {2-3i}2\right) & = 0 \\ x^2 - \left(\frac {2+3i}2 + \frac {2-3i}2\right)x + \left(\frac {2+3i}2\right)\left(\frac {2-3i}2\right) & = 0 \\ x^2 - 2x + \frac {13}4 & = 0 \\ \implies 4x^2 - 8x + 13 & = 0 \end{aligned}$

$[x-\dfrac{1}{2}(2+3i)][x-\dfrac{1}{2}(2-3i)]=0$

$(x-\dfrac{2}{2}-\dfrac{3i}{2})(x-\dfrac{2}{2}+\dfrac{3i}{2})=0$

$(x-1-\dfrac{3i}{2})(x-1+\dfrac{3i}{2})=0$

Multiplying both sides by $2$ , we get

$(2x-2-3i)(2x-2+3i)=0$

$[(2x-2)-3i][(2x-2)+3i]=0$

$(2x-2)^2-9i^2=0$

but: $i^2=-1$

$4x^2-8x+4-9(-1)=0$

$4x^2-8x+4+9$

$\color{#D61F06}\boxed{4x^2-8x+13=0}$