Question

How do I solve: x^2-14x+58=0

Answer 1

Compare your quadratic equation with \(ax^2+bx+c=0\)
find \[a=...?\\b=...?\\c=...?\\\] \[ \\ \sqrt{b^2-4ac}=...?\]
then the two roots of x are:
\(\huge{x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)