`x^2`
Solving the quadratic equation.
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)`