Chapter 2 - Linear and Quadratic Functions - Section 2.7 Complex Zeros of a Quadratic Function* - 2.7 Assess Your Understanding - Page 177: 3

$10-5i$

Multiple using the distributive property to obtain: \begin{align*} (2+i)(3-4i)&=2(3-4i)+i(3-4i)\\ &=2(3)+2(-4 i)+i(3)+i(-4 i)\\ &=6-8i+3 i-4 i^{2}\\ &=6-5i-4i^2\end{align*} Since $i^{2}=-1$, then \begin{align*} &=6-5i-4(-1)\\ &=6-5i+4\\ &=10-5i \end{align*}