# How do you simplify (6 + i )(8 – 3i)?

It is $51 - 10 i$
If you add or multiply complex numbers in algebraic form you do it as if they were real numbers but you have to remember that the square of imaginary unit $i$ equals to $- 1$
$\left(6 + i\right) \left(8 - 3 i\right) = 48 - 18 i + 8 i - 3 {i}^{2} = 48 - 18 i + 8 i + 3 = 51 - 10 i$