This is the equation for the solution of a second order polynomial of the form `aX^2+bX+C = 0` where the solution produces two roots
`(-b +- sqrt(b^2 -4ac))/(2a)`
Solving the quadratic equation.
Suppose `a x^2+b x+c=0` and `a!=0`
first divide by `a` to get:
`x^2+b/a x+c/a=0`
Then complete the square and obtain:
`x^2+b/a x+(b/(2a))^2-(b/(2a))^2+c/a=0`
The first three terms factor:
`(x+b/(2a))^2=(b^2)/(4a^2)-c/a`
Take square roots on both sides to get
`x+b/(2a)=+-sqrt((b^2)/(4a^2)-c/a)`
Finally move the b/(2a) to the right and simplify to get the two solutions:
`x_(1,2)=(-b+-sqrt(b^2-4a c))/(2a)`