Amazing.

Comparison with the form $Ax^2+Bx+C=0$ gives us $A=c,B=b,C=a$ . Therefore, by quadratic formula, $x_1 , x_2 = \frac{-B\pm \sqrt{B^2 - 4AC}}{2A}=\frac{-b\pm \sqrt{b^2-4ca}}{2c}=\boxed{\displaystyle{\frac{-b\pm \sqrt{b^2-4ac}}{2c}}}$ By switching the constant term and the leading coefficient of the quadratic equation, only the denominator of quadratic formula changes. The rest are essentially the same. :)

Here's my solution, I apologise for writing it completely in LaTex, but it made typing it so much easier!

$Starting\quad with\quad the\quad equation:\\ { cx }^{ 2 }+bx+a=0\\ \\ We\quad can\quad first\quad factorise\quad by\quad c:\\ c({ x }^{ 2 }+\frac { bx }{ c } +\frac { a }{ c } )\quad =\quad 0\\ Now,\quad completing\quad the\quad square\quad inside\quad the\quad brackets\quad gives:\\ c({ (x }+\frac { b }{ 2c } { ) }^{ 2 }-\frac { { b }^{ 2 } }{ { 4c }^{ 2 } } +\frac { a }{ c } )\quad =\quad 0\\ Multiplying\quad \frac { a }{ c } \quad by\quad \frac {4c}{4c}\quad will\quad give:\\ c({ (x }+\frac { b }{ 2c } { ) }^{ 2 }-\frac { { b }^{ 2 }-4ac }{ { 4c }^{ 2 } } )\quad =\quad 0\\ Multiplying\quad out\quad by\quad c\quad again\quad gives:\\ c{ (x }+\frac { b }{ 2c } { ) }^{ 2 }-\frac { { b }^{ 2 }-4ac }{ { 4c } } \quad =\quad 0\\ Taking\quad the\quad negative\quad to\quad the\quad right\quad gives:\\ c{ (x }+\frac { b }{ 2c } { ) }^{ 2 }\quad =\quad \frac { { b }^{ 2 }-4ac }{ { 4c } } \\ Dividing\quad by\quad c:\\ { (x }+\frac { b }{ 2c } { ) }^{ 2 }\quad =\quad \frac { { b }^{ 2 }-4ac }{ { 4{ c }^{ 2 } } } \\ Square\quad rooting\quad both\quad sides:\\ { x }+\frac { b }{ 2c } \quad =\quad \frac { \pm \sqrt { { b }^{ 2 }-4ac } }{ { { 2c } } } \\ And\quad finally\quad subtracting\quad \frac { b }{ 2c } \quad gives:\\ \boxed { x\quad =\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ { { 2c } } } }$