find the sum of 4a^3-7a-5 and -8a^2-2a+7 IS THE ANSWER

HOW TO do this The sum of four consecutive integers is at least 114. Find the smallest possible values for these numbers.

Originally Posted by Satnam
find the sum of 4a^3-7a-5 and -8a^2-2a+7 IS THE ANSWER
$(4a^3 - 7a - 5) + (-8a^2 - 2a + 7) = 4a^3 - 8a^2 - 9a + 2$
$n + (n+1) + (n+2) + (n+3) \ge 114$ solve for n