#### Answer

$\dfrac{3}{10}-\dfrac{8}{5}i$

#### Work Step by Step

To simplify the given expression, their is a need to rationalize the denominator. This can be done by multiplying the conjugate of the denominator, which is $2-4i$, to both the numerator and the denominator to obtain:
$\dfrac{7-2i}{2+4i}
\\=\dfrac{(7-2i)(2-4i)}{(2+4i)(2-4i)}$
RECALL:
(1) $(a-b)(a+b) = a^2-b^2$
(2) $(a+b)(c+d)=ac +ad + bc+bd$ (popularly known as the FOIL method)
Use the rule in (1) above to simplify the denominator and the FOIL method to expand the numerator to obtain:
$=\dfrac{7(2)+7(-4i)+(-2i)(2)+(-2i)(-4i)}{2^2-(4i)^2}
\\=\dfrac{14-28i-4i+8i^2}{4-16i^2}
\\=\dfrac{14-32i+8i^2}{4-16i^2}$
Since $i^2=1$, the expression above simplifies to:
$=\dfrac{14-32i+8(-1)}{4-16(-1)}
\\=\dfrac{14-32i-8}{4+16}
\\=\dfrac{6-32i}{20}
\\=\dfrac{2(3-16i)}{2(10)}$
Simplify by canceling the common factor $2$ to obtain:
$=\dfrac{3-16i}{10}
\\=\dfrac{3}{10} - \dfrac{16}{10}i
\\=\dfrac{3}{10}-\dfrac{8}{5}i$