(20+2x)(40-x)=1200

问题描述:

(20+2x)(40-x)=1200

(20+2x)(40-x)=1200
800+80x-20x-2x^2=1200
2x^2-60x+400=0
x^2-30x+200=0
(x-15)^2=25
x-15=5或x-15=-5
x=20或x=10