Question 1013184: Hi, can you help me figure out this equation: Find the zeros of ax^2+bx=y-c by completing the square.
Answer by fractalier(6550)   (Show Source):
You can put this solution on YOUR website! From another website comes this...
This is the original equation. ax^2 + bx + c = 0
Move the loose number to the other side. ax^2 + bx = –c
Divide through by whatever is multiplied on the squared term.
Take half of the x-term, and square it. Add the squared term to both sides.
x^2 + (b/a)x + (b^2/4a^2) = –(c/a) + (b^2/4a^2)
Simplify on the right-hand side; in this case, simplify by converting to a common denominator.
x^2 + (b/a)x + (b^2/4a^2) = –(4ac/4a^2) + (b^2/4a^2)
Convert the left-hand side to square form (and do a bit more simplifying on the right).
(x + b/2a)^2 = (b^2 – 4ac)/4a^2
Square-root both sides, remembering to put the "±" on the right.
x + b/2a = ± sqrt(b^2 – 4ac)/2a
Solve for "x =", and simplify as necessary.
x = [ –b ± sqrt(b^2 – 4ac) ] / 2a
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