No, differentiation is Calc, this is only algebra. You derive the quad. form. by completing the square. Like this:
ax^2 +bx +c =0
x^2 +(bx)/a +c/a =0
x^2 + (bx)/a = -c/a
x^2 +(bx)/a +[(b^2)/4a^2]* = (b^2)/4a^2 - c/a
[x+(b/2a)]^2 =+ sqrt[(b^2 - 4ac)/4a^2]
x= [-b +sqrt(b^2 -4ac)]/2a
*The generic form for the sum of two squares is x^2 +2yx +y^2. At this point, I consolidated a few things. Since x^2 + (bx)/a is equal to x^2 +2[(b)/2a]x, it follows that in this case, y= b/2a. We square that and insert it here and on the other side for balance.
#It wasn't marked, but someone may be confused as to what sqrt is. It's a contraction of square root, and applies to the term in the groupers following it.