Let's solve the equation by completing the square:
[tex]\begin{gathered} a^2+2a-3=0 \\ a^2+2a=3 \\ a^2+2a+(\frac{2}{2})^2=3+(\frac{2}{2})^2 \\ a^2+2a+1^2=3+1^2 \\ (a+1)^2=4 \\ \sqrt[]{(a+1)^2}=\sqrt[]{4} \\ \lvert a+1\rvert=2 \\ a+1=\pm2 \\ \text{then} \\ a+1=2 \\ a=2-1 \\ a=1 \\ or \\ a+1=-2 \\ a=-2-1 \\ a=-3 \end{gathered}[/tex]therefore the solutions are a=1 and a=-3
