how do I write (8-3i)(-2-3i) in standard form?

FOIL the binomials

-16 - 24i + 6i + 9i^2 ... i^2 is (√-1)^2

collect like terms

-25 - 18i

thank you so much

again, thank you, but if I may ask, how did you get the square root sign? I am just trying to understand it:)

To write the expression (8-3i)(-2-3i) in standard form, follow these steps:

Step 1: Simplify the given expression by using the distributive property to multiply the two binomials together.
(8-3i)(-2-3i) = 8(-2) + 8(-3i) - 3i(-2) - 3i(-3i)

Step 2: Simplify further by performing the multiplication.
= -16 - 24i + 6i + 9i^2

Step 3: Replace i^2 with -1, as i^2 is defined as -1.
= -16 - 24i + 6i + 9(-1)

Step 4: Continue simplifying.
= -16 - 24i + 6i - 9

Step 5: Combine like terms.
= -25 - 18i

Therefore, the expression (8-3i)(-2-3i) simplifies to -25 - 18i in standard form.