$1)\,{x^2} + 2x - 3 = 0 \Rightarrow $
$\Delta  = {b^2} - 4ac = {(2)^2} - 4(1)( - 3) = 16$
${x_{1,2}} = \frac{{ - b \pm \sqrt \Delta  }}{{2a}} = \frac{{ - 2 \pm \sqrt {16} }}{2}$
$ \Rightarrow \left\{ {\begin{array}{*{20}{c}}
{{x_1} = \frac{{ - 2 + 4}}{2} = \frac{2}{2} = 1\,\,\,\,\,\,}\\
{{x_2} = \frac{{ - 2 - 4}}{2} = \frac{{ - 6}}{2} =  - 3}
\end{array}} \right.$

$2)\,{x^2} - 2x + 3 = 0 \Rightarrow $
$\Delta  = {b^2} - 4ac = {( - 2)^2} - 4(1)(3) = 4 - 12 =  - 8$
$3)\,{x^2} + 2x + 3 = 0 \Rightarrow $
$\Delta  = {b^2} - 4ac = {(2)^2} - 4(1)(3) = 4 - 12 =  - 8$
$4)\,{x^2} - 2x - 3 = 0 \Rightarrow $
$\Delta  = {b^2} - 4ac = {( - 2)^2} - 4(1)( - 3) = 4 + 12 = 16$
${x_{1,2}} = \frac{{ - b \pm \sqrt \Delta  }}{{2a}} = \frac{{2 \pm 4}}{{2 \times 1}} = \left\{ {\begin{array}{*{20}{c}}
{\frac{{2 + 4}}{2} = \frac{6}{2} = 3\,\,\,\,\,}\\
{\frac{{2 - 4}}{2} = \frac{{ - 2}}{2} =  - 1}
\end{array}} \right.$