Let's solve the equation:
[tex]\begin{gathered} x^2+4x-5=0 \\ x^2+4x=5 \\ x^2+4x+(\frac{4}{2})^2=5+(\frac{4}{2})^2 \\ (x+2)^2=5+\frac{16}{4} \\ (x+2)^2=\frac{36}{4} \\ x+2=\pm\sqrt[]{\frac{36}{4}} \\ x+2=\pm3 \\ \text{ then} \\ x=-2+3 \\ x=1 \\ or\text{ } \\ x=-2-3 \\ x=-5 \end{gathered}[/tex]
Therefore, the solutions of the equation are x=1 and x=-5
