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We first divide by a to get $%num=1% x^2+b/ax+c/a=0$

Then we complete the square and obtain $%num% x^2+b/ax+(b/(2a))^2-(b/(2a))^2+c/a=0$

The first three terms factor to give $%num% (x+b/(2a))^2=(b^2)/(4a^2)-c/a$

Now we take square roots on both sides and get $%num% x+b/(2a)=+-sqrt((b^2)/(4a^2)-c/a)$

Finally we move $b/(2a)$ to the right and simplify to get the two solutions: $%num=5% x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)$

Adding a silly note to get a quote as a superscript r' vs. r'