Solve: (2i+1)(3-i)...

I know the answer is 5+5i, but could you show me the steps on how to get that answer? Thanks! (:

(2i+1)(3-i) = 6i-2i²+3-i

i²=-1 so:

6i-2i²+3-i = 6i-2(-1)+3-i =6i+2+3-i = 5+5i

(2i + 1)(3 - i)
= - (2i + 1)(i - 3)
= - (2i² - 6i + i - 3)
= - {2(- 1) - 5i - 3}
= - {- 2 - 5i - 3}
= - (- 5 - 5i)
= 5 + 5i

(2i+1)(3-i) = 6i - 2i^2 +3 - i =

= 6i + 2 + 3 - i = 5 + 5i (remember that i^2 = -1

(2i + 1)(3 - i)

= 6i - 2i^2 + 3 - i

= 5i - 2(-1) + 3

= 5i + 5

(2i+1)(3-i) = (2i)(3)+(2i)(-i)+(1)(3)+(1)(-i)

= 6i -2i^2 +3 -i = 6i +2 +3 -i = 5 + 5i

(2i+1)(3-i)=
2i*3-2i*i+1*3-1i=
6i-2i^2+3+1i=
6i+2+3-1i=
5i+5=
5+5i

F.O.I.L. first, outside, inside, last