Question 1076984
Find the zeroes of ax^2+bx = y-c 
By completing the square 
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y = ax^2 + bx + c = 0
Divide by a
x^2 + (b/a)x + c/a = 0
x^2 + (b/a)x = -c/a
Make left side a square by adding (b/2a)^2
x^2 + (b/a)x + b^2/4a^2 = -c/a + b^2/4a^2
{{{(x + b/2a)^2 = b^2/4a^2 - 4ac/4a^2}}}
{{{(x + b/2a)^2 = (b^2 - 4ac)/4a^2}}}
{{{x + b/2a = sqrt((b^2 - 4ac)/4a^2)}}}
{{{x + b/2a = sqrt(b^2 - 4ac)/2a}}}
{{{x = -b/2a + sqrt(b^2 - 4ac)/2a}}}
{{{x = (-b +- sqrt(b^2 - 4ac))/2a}}}