Question

5 years ago, a womens age was the square of her son's age ,10 years hence, her age will be twice that of her sons age find the present age of the women

Answer 1

five years ago a woman age was the square of her son age.
ten years hence the mother age was twice of his son age.
:
Let m = mother's age
Let s = son's age
:
m - 5 = (s-5)^2
m - 5 = s^2 - 10s + 25
m = s^2 - 10s + 25 + 5
m = s^2 - 10s + 30
:
We also know,"ten years hence the mother age was twice of his son age"
m + 10 = 2(s+10)

m + 10 = 2s + 20
m = 2s + 20 - 10
m = 2s + 10
therefore
s^2 - 10s + 30 = 2s + 10
s^2 - 10s - 2s + 30 - 10 = 0
A quadratic equation
s^2 - 12s + 20 = 0; Corrected a mistake here, Sorry!
Factors to
(s-10)(s-2) = 0
The only solution that will make sense here. 
s = 10 yrs old is son's age

Answer 2

Let the present age of the son 5 years ago=x years.
The woman's age 5 years ago= x^{2}
The present age of the woman=x^{2}  +5 years
The present age of her son age=x+5 years
10 years hence:
the woman's age will be=(x^{2})+10 years +x^{2}  +15 years
her son age will be=(x+5)+10 years=x+15
according to the given statement:
10 years hence, woman's age=twice her son's age
x^{2}  +15 years=2(x+15)
x^2x-15=0
on solving we get x=5 or -3
age cannot be negative so x=5
Present age of woman=x^+5=5*5+5=30 years