If (x+a)(x+b) = (x+c)(x+d) then x=___

I'm reviewing for an entrance test. I'm posting only the questions that I couldn't figure out on my own.

Foil both sides:
x^2+ax+bx+ab = x^2+cx+dx+cd

The x^2 on both sides cancel:
ax+bx+ab = cx+dx+cd

Move all the terms with x to one side, and everything else to the other side:
ax+bx-cx-dx = cd-ab

Factor out the x:
x(a+b-c-d) = cd-ab

Divide both sides by (a+b-c-d) to get the answer:
x= (cd-ab) / (a+b-c-d)

For (x+a)(x+b) We have,.,,
1)
x + a = (x + c)(x + d)
x = (x + c)(x + d) - a

2)
x+b = (x + c)(x + d)
x = (x + c)(x + d) - b

Therefore,
x = (x + c)(x + d) - a OR x = (x + c)(x + d) - b
Remember that this is a square, hence its degree is two, thus two possible solutions =)

x^2 + ax + bx + ab - x^2 - cx - dx - cd = 0
ax + bx - cx - dx = cd - ab
x(a + b - c - d) = cd - ab
x = (cd-ab)/(a+b-c-d)

x^2+(a+b)x+ab=x^2+(c+d)x+cd
(a+b-c-d)x=cd-ab
x=(cd-ab)/(a+b-c-c)
God bless you.