How do you perform the operation and write the result in standard form given (1+i)(3-2i)?

$5 + i$
We know that ${i}^{2} = - 1$
$\left(1 + i\right) \left(3 - 2 i\right) = 3 - 2 i + 3 i - 2 {i}^{2} = 3 + i - 2 \left(- 1\right) = 5 + i$