Question 627671
Let
Present age of woman = w years old
Present age of daughter = d years old


woman is 4 times older than her daughter
w=4d............(1)
6 years ago 
woman was = w-6 years old
daughter was = d-6
then
product of their ages was 136
(w-6)(d-6)=136
substitute the value of from (1) to above equation
(4d-6)(d-6)=136
4d^2-24d-6d+36=136
4d^2-30d+36=136
2(2d^2-15d+18)=136
Divide by 2 both sides
2(2d^2-15d+18)/2=136/2
2d^2-15d+18=68
2d^2-15d+18-68=0
2d^2-15d-50=0
*[invoke quadratic "d", 2, -15, -50 ]
d=10 or d=-2.5 (unacceptable)
so
d=10
Put the value of d in (1)
w=4d
w=4(10)
w=40


Present age of woman = w = 40 years old
Present age of daughter = d = 10 years old