Suppose ax^2+bx+c=0 and a!=0. We first divide by a to get x^2+b/ax+c/a=0.

Then we complete the square and obtain x^2+b/ax+(b/(2a))^2-(b/(2a))^2+c/a=0.
The first three terms factor to give (x+b/(2a))^2=(b^2)/(4a^2)-c/a.
Now we take square roots on both sides and get x+b/(2a)=+-sqrt((b^2)/(4a^2)-c/a).

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

Comment

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• sin^-1(x) function names are treated as constants
• d/dxf(x)=lim_(h->0)(f(x+h)-f(x))/h complex subscripts are bracketed, displayed under lim
• $\frac{d}{dx}f(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ standard LaTeX notation is an alternative
• f(x)=sum_(n=0)^oo(f^((n))(a))/(n!)(x-a)^n must be bracketed, else the numerator is only a  f(x)=\sum_{n=0}^\inft \frac{f^{(n)}(a)}{n!}(x-a)^nstandard LaTeX produces the same result int_0^1f(x)dx subscripts must come before superscripts
• [[a,b],[c,d]]((n),(k)) [abcd](nk) matrices and column vectors are simple to type
• x/x={(1,if x!=0),(text{undefined},if x=0):} x/x={(1,if x!=0),(text{undefined},if x=0):} piecewise defined function are based on matrix notation a//b a/b use // for inline fractions
• (a/b)/(c/d) (a/b)/(c/d) with brackets, multiple fraction work as expected a/b/c/d ab/cd without brackets the parser chooses this particular expression ((a*b))/conly one level of brackets is removed; * gives standard product
• sqrtsqrtroot3x spaces are optional, only serve to split strings that should not match
• (:a,b:) and x lt y lt 1