Question: So i want 12.6666667 to show up as 12.6 and not 13 or 12.67
This is what i have done

```
Math.Round((5 * (53 - 32)) / 9, 1)

```

Answer: 
`Math.Truncate (x * 10.0) / 10.0` is one way to do it. 

Using your numbers: 

```
double x = 5.0 * (53.0 - 32.0) / 9.0;
double result = Math.Truncate (x * 10.0) / 10.0;

```

Note the use of ".0" at the end of all numbers, this ensures that floating-point math, and not integer math, is used when combining them. Integer math will remove any remainders when dividing numbers, while floating point will keep the stuff to the right of the decimal point, which is what you want.
If you want 12.6666667 to show up as 12.6, you'll have to truncate, not round; otherwise you'd get 12.7 to one decimal place.

Personally I'd do that by going ((int) (12.6666667 \* 10)) / 10.