Presumably you're solving for y.
\(0=(2y-1)\times(8-y)\)
You have two chunks of numbers (each set of parentheses is one chunk), two values, which when multiplied equal 0. The only way this equation can make sense is if either
\((2y-1)=0\)
or else
\((8-y)=0\)
If neither of these values is 0, then there is no way to multiply the two values and get 0 as your answer. Therefore, there are two possible solutions to this equation. Either (2y-1)=0 or (8-y)=0
So you solve each of those two equations out for y, and that will give you the two values that y could equal.
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