solving formula

for ax^2+bx+c=0

where does it come from?

ax^2 + bx + c = 0
x(ax+b) + c = 0
x(ax+b) = -c
...
aaaaargh

http://en.wikipedia.org/wiki/Quadratic_equation#Derivation

UNDERAGE BAMPERSAND.

I can't believe this fagot doesn't know quadratic.

ax^2+bx+c=0
ax^2+bx=-c
x^2+(bx/a)=-c/a
x^2+(bx/a)+[(b^2)/(4a^2)]=-c/a + [(b^2)/(4a^2)]
(x+ b/2a)^2= (-4ac/(4a^2)) + [(b^2)/(4a^2)]
(x+ b/2a)^2= (b^2-4ac)/(4a^2)
x+ b/2a= (+ or -) (sqrt(b^2-4ac))/2a
x= -b/2a (+ or -) (sqrt(b^2-4ac))/2a
x= [-b (+ or -) (sqrt(b^2-4ac))]/2a

Learn to complete the square.

First note that

(2ax + b)² = (2ax)² + 4abx + b²
(2ax)² + 4abx = (2ax + 4b)² - (4b)²  [*]

Then,

ax² + bx + c = 0
[multiply by 4a for prettier numbers]
(2ax)² + 4abx + 4ac = 0
[using *]
(2ax + b)² - b² + 4ac = 0

and thus:
(2ax + b)² = b² - 4ac

etc..

#6, I like your version better, since you append comments to explain WHY you performed the operation.  Give yourself a gold star.  Too often we just spurt strings of operations at each other in some sort of mathematical bukkake, but the meaning is lost.

The word pirahna, is all I can think of that rhymes with marijuana

Marijuana MUST be legalized.