Solution. Factorising \begin {align*} x^2+3x+2x+6 & = 0\\ x(x+3)+2(x+3) & =0\\ \implies (x+3)(x+2) & = 0\\\\ \implies \text {either}\quad x+3=0\quad &\text {or}\quad x+2=0\\ \implies \text {either}\quad x=-3\quad & \text {or}\quad x=-2\\ \end {align*}

The other method of solving a quadratic equation is by completing the squares. To do this observe that \((x+p)^2=x^2+2px+p^2\). But \[p^2=\left [\frac {1}{2}(2p)\right ]^2\] In other words a complete square is such that the last term \(p^2\) is the square of \(\frac {1}{2}\) the middle part \(2p\).

Our quadratic equation has the form \(ax^2+bx+c=0\)

To complete the square write \(ax^2+bx=-c\) \[\implies x^2+\frac {b}{a}x=-\frac {c}{a}\] middle constant is \(\frac {b}{a}\) half of \(\frac {b}{a}\) is \(\frac {b}{2a}\). \begin {align*} \implies x^2+\frac {b}{a}x+\left (\frac {b}{2a}\right )^2 & =\frac {-c}{a}+\left (\frac {b}{2a}\right )^2\\\\ \implies \left (x+\frac {b}{2a}\right )^2 & = \frac {-c}{a}+\frac {b^2}{4a^2}\\\\ \implies \left (x+\frac {b}{2a}\right )^2 & =\frac {-4ac+b^2}{4a^2} \end {align*}

Take the square root. \begin {align*} x+\frac {b}{2a} & = \pm \sqrt {\frac {b^2-4ac}{4a^2}}\\ \implies x & = \frac {-b}{2a}\pm \frac {\sqrt {b^2-4ac}}{\sqrt {4a^2}}\\ & = \frac {-b}{2a} \pm \frac {\sqrt {b^2-4ac}}{2a}\\ \implies x & = \frac {-b\pm \sqrt {b^2-4ac}}{2a}\\\\ \text {i.e either}\quad x &=\frac {-b+\sqrt {b^2-4ac}}{2a}\quad \text {or} \quad x=\frac {-b-\sqrt {b^2-4ac}}{2a}\\ \end {align*}