How can i do this? i don't know how to answer this cause fractions..

If    $x=\,-\,\frac83$   , does    $2(3x^2+24x+48)+1(x-6)=2(1)$   ?

Replace every   " x "   with   $-\,\frac83$   in the second equation.

$2(3(-\frac83)^2+24(-\frac83)+48)+1((-\frac83)-6)\stackrel{?}{=}2(1) \\~\\ 2(3(\frac{64}{9})-64+48)+(-\frac{26}3)\stackrel{?}{=}2 \\~\\ 2(\frac{64}{3}-16)-\frac{26}3\stackrel{?}{=}2 \\~\\ 2(\frac{16}{3})-\frac{26}3\stackrel{?}{=}2 \\~\\ \frac{32}{3}-\frac{26}3\stackrel{?}{=}2 \\~\\ \frac63\stackrel{?}{=}2 \\~\\ 2=2\quad\text{ true!}$