Find the product:

By horizontal method:

$\left(2 x^{2}+3 x-7\right) \times\left(3 x^{2}-5 x+4\right)$

$=2 x^{2}\left(3 x^{2}-5 x+4\right)+3 x\left(3 x^{2}-5 x+4\right)-7\left(3 x^{2}-5 x+4\right)$

$=6 x^{4}-10 x^{3}+8 x^{2}+9 x^{3}-15 x^{2}+12 x-21 x^{2}+35 x-28$

$=6 x^{4}-x^{3}-28 x^{2}+47 x-28$