f.4 Quadratic Question

(x+1)^2+3(x+1)-4=0
Let y=x+1
y^2+3y-4=0
(y+4)(y-1)=0
y+4=0 or y-1=0
y=-4 or y=1
x+1=-4 or x+1=1
x=-5 or x=0

(4-x)^2-5(4-x)-14=0
[(4-x)+2][(4-x)-7]=0
(6-x)(-3-x)=0
(x+3)(x-6)=0
x+3=0 or x-6=0
x=-3 or x=6


2011-09-14 20:02:56 補充：
Yes, I agree with "知識就是力量".

There is another approach as follows:

(x+1)^2+3(x+1)-4=0
x^2+2x+1+3x+3-4=0
x^2+5x=0
x(x+5)=0
x=-5 or x=0

2011-09-14 20:05:44 補充：
(4-x)^2-5(4-x)-14=0
16-8x+x^2-20+5x-14=0
x^2-3x-18=0
(x+3)(x-6)=0
x+3=0 or x-6=0
x=-3 or x=6

Another approach:
Expand all the complete squares and regroup them into a new quadratic equation, but it is a more complicated way.