Answer
$y^{2}-19y+16$
Work Step by Step
Subtract the polynomials inside the first brackets
$ \begin{array}{llll}
& 4y^{2} & -3y & +8 \\
-( & 5y^{2} & +7y & -4)\\
\hline & & & \\
= & -y^{2} & -10y & +12\\
& & &
\end{array} $
Add the polynomials in the second brackets:
$ \begin{array}{llll}
& 8y^{2} & +5y & -7 \\
+( & -10y^{2} & +4y & +3)\\
\hline & & & \\
= & -2y^{2} & +9y & -4
\end{array} $
Now subtract the results of the brackets:
$ \begin{array}{llll}
& -y^{2} & -10y & +12 \\
-( & -2y^{2} & +9y & -4)\\
\hline & & & \\
= & y^{2} & -19y & +16
\end{array} $