Expand `(x+5-2i)(x+5+2i)`

We use the extended distributive property:

`(x+5-2i)(x+5+2i)`

`=x^2+5x+2ix+5x+25+10i-2ix-10i-4i^2`

`=x^2+10x+25-4i^2` (but `i=sqrt(-1)=>i^2=-1` so)

`=x^2+10x+29`

**Thus** `(x+5-2i)(x+5+2i)=x^2+10x+29`

Expand `(x+5-2i)(x+5+2i)`

We use the extended distributive property:

`(x+5-2i)(x+5+2i)`

`=x^2+5x+2ix+5x+25+10i-2ix-10i-4i^2`

`=x^2+10x+25-4i^2` (but `i=sqrt(-1)=>i^2=-1` so)

`=x^2+10x+29`

**Thus** `(x+5-2i)(x+5+2i)=x^2+10x+29`