If f(x) = x^3 + 2x^2-29x-30 and f(-6) = 0, then find all of the zeros of f(x) algebraically

there are going to be three zeros.

x=-1

x=-6

x=5

let x and 400 - x be the two numbers.  

==> 0.80x + 0.85(400 - x) = 400 - 68 from the statement of the problem.  

< ==> 0.80x + 0.85(400 - x) = 400 - 68  

==> 340 - 0.05x = 332  

< ==> -0.05x = -8 ==> x = 160, the first number ==> 400 - x = 240, the second number.