Question

If x a set of all real numbers R the find the solution set for $(x^2-2x-35)<0$

Answer 1

(x^2-2x-35)<0
(x+5)(x-7)<0

Case 1 :
When (x+5) < 0 and (x-7) >0
i.e. x < -5 and x > 7
This is not possible as as a number less than -5 cannot be greater than 7

Case 2:
When (x+5) > 0 and (x-7) < 0
i.e. x > -5 and x < 7
-5 < x < 7, which is possible.
Therefore Solution set = $(-5,7)$