1)
We note that the quadratic can be factored into [tex](2x-1)(x-1)[/tex].
The quadratic is greater than [tex]0[/tex] if both of its factors are positive, or they are both negative.
Case: both positive
We need to solve the system of inequalities:
[tex]2x-1>0\\x-1>0[/tex].
The first inequality gives [tex]x>\frac{1}{2}[/tex].
The second inequality gives [tex]x>1[/tex].
Taking the points where the inequalities coincide gives [tex]x>1[/tex].
(Note: 1 is a root of the quadratic. Coincidence? If not, try and prove it!)
Case: both negative
We need to solve the system of inequalities:
[tex]2x-1<0\\x-1<0[/tex].
The first inequality gives [tex]x<\frac{1}{2}[/tex].
The second inequality gives [tex]x<1[/tex].
Taking the points where the inequalities coincide gives [tex]x<\frac{1}{2}[/tex].
(Note: [tex]\frac{1}{2}[/tex] is the other root of the quadratic. Coincidence? If not, try and prove it!)
Taking the union of both cases gives the solution set: [tex]\boxed{x\in (-\infty, \frac{1}{2}) \cup (1, \infty)}[/tex]
2)
We bring over the [tex]15[/tex] to get [tex]x^2-2x-15<0[/tex].
Note that the quadratic factors into [tex](x-5)(x+3)[/tex].
The quadratic is less than [tex]0[/tex] if 1 of its factors is negative, but not both.
Case: first factor is negative, second positive
We have that [tex]x-5<0[/tex] and [tex]x+3>0[/tex].
We get that [tex]x<5[/tex] and [tex]x>-3[/tex], which has the solution set [tex]-3<x<5[/tex].
Case: second factor is negative, first positive
We have that [tex]x+3<0[/tex] and [tex]x-5>0[/tex].
We get that [tex]x<-3[/tex] and [tex]x>5[/tex], which has no solutions.
So, the solution set is [tex]\boxed{x\in (-3, 5)}[/tex]
