To find the product you have to multiply each term in the one parenthesis with each term in the second parenthesis and add together.
In this instance you will first multiply all the terms in $(3{a}^{2}+4a-4)$ with 2a, there after all the terms in $(3{a}^{2}+4a-4)$ with 9 and then add together
1. $(3{a}^{2}+4a-4)\cdot 2a=6{a}^{3}+8{a}^{2}-8a$
2. $(3{a}^{2}+4a-4)\cdot 9=27{a}^{2}+36a-36$
But as 9 is negative, "adding together" implies that the 2nd expression must be subtracted from the 1st:
$(6{a}^{3}+8{a}^{2}-8a)-(27{a}^{2}+36a-36)$ $=6{a}^{3}+(8-27){a}^{2}-(8+36)a+36$
(remember that -(-36)=+36)
$=6{a}^{3}-19{a}^{2}-44a+36$