#### Answer

$\dfrac{13-34i}{25}$

#### Work Step by Step

Multiplying by the conjugate of the denominator results to
\begin{array}{l}\require{cancel}
\dfrac{7-2i}{3+4i}
\\\\=
\dfrac{7-2i}{3+4i}\cdot\dfrac{3-4i}{3-4i}
\\\\=
\dfrac{(7-2i)(3-4i)}{(3+4i)(3-4i)}
.\end{array}
Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent
\begin{array}{l}\require{cancel}
\dfrac{(7-2i)(3-4i)}{(3+4i)(3-4i)}
\\\\=
\dfrac{(7-2i)(3-4i)}{(3)^2-(4i)^2}
\\\\=
\dfrac{(7-2i)(3-4i)}{9-16i^2}
.\end{array}
Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{(7-2i)(3-4i)}{9-16i^2}
\\\\=
\dfrac{7(3)+7(-4i)-2i(3)-2i(-4i)}{9-16i^2}
\\\\=
\dfrac{21-28i-6i+8i^2}{9-16i^2}
\\\\=
\dfrac{21-34i+8i^2}{9-16i^2}
.\end{array}
Since $i^2=-1,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{21-34i+8i^2}{9-16i^2}
\\\\=
\dfrac{21-34i+8(-1)}{9-16(-1)}
\\\\=
\dfrac{21-34i-8}{9+16}
\\\\=
\dfrac{13-34i}{25}
.\end{array}