Question 409110
SOLUTIONS: Let the three consecutive odd integers are x,(x+2),(x+4)

Given:x+(x+4)^2 = 236

x+ x^2 +8x +16 = 236

x^2 +9x - 220 =0    

By solving the quadratic equation

x^2 + 20 x -11x -220 =0   

x(x+20) -11(x+20)=0
(x+20)(x-11)=0
x+20=0
and x-11=0

so x=-20 and x = 11

so the three consecutive integers are 11. 11+2, 11+4

11,13,15