#### Answer

$2-5i$

#### Work Step by Step

Step 1: Multiplying both the numerator and the denominator by the complex conjugate of the denominator:
$\frac{25-19i}{5+3i}\times\frac{5-3i}{5-3i}$
Step 2: $\frac{(25-19i)(5-3i)}{(5+3i)(5-3i)}=\frac{25(5-3i)-19i(5-3i)}{(5)^{2}-(3i)^{2}}=\frac{125-75i-95i+57i^{2}}{25-9i^{2}}$
Step 3: $\frac{125-75i-95i+57i^{2}}{25-9i^{2}}=\frac{125-170i-57}{25+9}=\frac{68-170i}{34}$
Step 4: $\frac{68-170i}{34}=\frac{34(2-5i)}{34}=2-5i$