#### Answer

$-30-40i$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To simplify the given expression, $
-5i(3-i)^2
,$ use the special product on squaring binomials and the Distributive Property. Then use the equivalence $i^2=-1.$ Finally, combine combine the real parts and the imaginary parts.
$\bf{\text{Solution Details:}}$
Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
-5i[(3)^2-2(3)(i)+(i)^2]
\\\\=
-5i[9-6i+i^2]
.\end{array}
Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
-5i(9)-5i(-6i)-5i(i^2)
\\\\=
-45i+30i^2-5i(i^2)
.\end{array}
Since $i^2=-1,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
-45i+30(-1)-5i(-1)
\\\\=
-45i-30+5i
.\end{array}
Combining the real parts and the imaginary parts results to
\begin{array}{l}\require{cancel}
-30+(-45i+5i)
\\\\=
-30-40i
.\end{array}