Question 5912
I answered you incorrectly.  I read this as {{{25x^2+16=40x}}} rather than {{{25x^2+16<40x}}}

To solve this problem:

25x^2+16<40x
set this as an equal sign:
25x^2+16=40x
subtract 40x from both sides:
25x^2-40x+16=0
(5x-4)(5x-4)=0
x=4/5  You have two choices:  your solution set can be from negative infinity to 4/5 or from 4/5 to infinity:

Test using the numbers 3/5 (if the solution set is less than 4/5 then this number will be included)  Make a second test with the number 1 (if the solution set is from 4/5 to infinity, then the number 1 will be included in the set):

If we put 1 back into the equation:
25(1)^2+16<40(1)
25(1)+16<40
41<40  THIS IS NOT A TRUE STATEMENT, SO THE SOLUTION SET IS NOT GREATER THAN 4/5.

25(3/5)^2+16<40(3/5)
25(9/25)+16<24
9+16<24
25<24.  This is not a true statement so the solution set cannot be less than 4/5 either

so there is not a solution to this problem.