32x+3x+1-28<0,
32x+3x+1<28,
32x+3x+1<27+1,
32x+3x+1<33+30.
From this we would expect to solve by solving 2x=3 and x+1=0. However, the solutions of these two equations don't match, because we get x=1.5 and x=-1.
If we had, for example, 32x+1+3x-1<33+30, then 2x+1=3, so x=1; and x-1=0, so x=1. These results are consistent, so the solution would have been x<1, because of the inequality.
Or, another example, 32x+3x+1-18<0, ..., 32x+3x+1<32+32, so 2x=2 and x+1=2 both give x=1 as solutions. Therefore, x<1, because of the inequality. I believe that 32x+3x+1-18<0 was the intended question.
x<1.261860 approx assuming the question has been correctly stated. The method of solution involves use of a calculator. First, we note that when x=1, we get:
32+32=9+9=18 and 18 is less than 28. So the inequality is true.
When x=2: 34+33=81+27=108 which exceeds 28, so the inequality is false, and x must be between 1 and 2. So, using the calculator we set x=1.5. 33+32.5=42.59 approx which exceeds 28. Therefore, x is between 1 and 1.5, so we next set x=1.25: 32.5+32.25=27.43 approx which is less than 28. Therefore, x is between 1.25 and 1.5, and we next set x=1.375, and so on, getting closer to a solution. The solution is irrational, so we continue until we reach some acceptable degree of accuracy, hence x<1.261860 to 6 decimal places. There are other methods of solving this, usually involving iteration.