What is the standard form of (5+i) (6-5i) / 2i (-3+7i) ?

Since all parts of 2i (-3+7i) is in the denominator, you have to use an extra set of parentheses around the whole denominator. this is what you do

(5+i) (6-5i) / (2i (-3+7i))

= (30-25i+6i-5i^2) / (-6i+14i^2)

= (35-19i) / (-14-6i)

= (35-19i) (-7+3i) / (2 (-7-3i) (-7+3i))

= (-245+105i+133i-57i^2) / (2 (49-9i^2))

= (-188+238i) / (2 (58))

= (-188+238i) / 116

= - 47/29 + 119/58 i so that is how you do it

(35-19i) / (-14-6i)

Foil the numerator. Distribute the denominator.