Solusi

$ax^2+bx+c=0$

$\large{x^2+\frac{b}{a}x+\frac{c}{a}=0}$

$\large{x^2+\frac{b}{a}x=-\frac{c}{a}}$

$\large{x^2+2.\frac{b}{2a}x=-\frac{c}{a}}$

$\large{x^2+2.\frac{b}{2a}x+(\frac{b}{2a})^2=(\frac{b}{2a})^2-\frac{c}{a}}$

$\large{(x + \frac{b}{2a})^2=\frac{b^2}{4a^2}-\frac{c}{a}}$

$\large{(x + \frac{b}{2a})^2=\frac{b^2}{4a^2}-\frac{4ac}{4a^2}}$

$\large{(x + \frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}}$

$\large{|x + \frac{b}{2a}|=\frac{\sqrt{b^2-4ac}}{2a}}$

Kemungkinan 1: $x + \frac{b}{2a} \lt 0$

$\large{x + \frac{b}{2a}=-\frac{\sqrt{b^2-4ac}}{2a}}$

$\large{x =-\frac{b}{2a}-\frac{\sqrt{b^2-4ac}}{2a}}$

$\large{x =\frac{-b-\sqrt{b^2-4ac}}{2a}}$

Kemungkinan 2: $x + \frac{b}{2a} \ge 0$

$\large{x + \frac{b}{2a}=\frac{\sqrt{b^2-4ac}}{2a}}$

$\large{x =-\frac{b}{2a}\frac{\sqrt{b^2-4ac}}{2a}}$

$\large{x =\frac{-b+\sqrt{b^2-4ac}}{2a}}$

Gabungkan kedua kemungkinan

$\large{x =\frac{-b \pm \sqrt{b^2-4ac}}{2a}}$

Referensi
Solusi ini menggunakan teknik yang sama dengan dengan solusi soal 18 (solusi 2) dan soal 19 (solusi 2)