Given the quadratic equation:
[tex]x^2+12x=13[/tex]We can rewrite the equation as follows:
[tex]\begin{gathered} x^2+12x-13=0 \\ x^2+2\cdot6\cdot x-13=0 \\ x^2+2\cdot6\cdot x+6^2-6^2-13=0 \\ (x^2+2\cdot6\cdot x+6^2)-36-13=0 \end{gathered}[/tex]We see that the term inside the parenthesis is a perfect square polynomial. Then:
[tex](x+6)^2-49=0[/tex]Solving for x:
[tex](x+6)^2=49[/tex]Taking the square on both sides:
[tex]\begin{gathered} \sqrt{(x+6)^2}=\sqrt{49} \\ \\ |x+6|=7 \end{gathered}[/tex]This equation can turn into two equations:
[tex]\begin{gathered} x+6=7...(1) \\ x+6=-7...(2) \end{gathered}[/tex]Solving (1):
[tex]\begin{gathered} x=7-6 \\ \\ \Rightarrow x=1 \end{gathered}[/tex]Solving (2):
[tex]\begin{gathered} x=-7-6 \\ \\ \Rightarrow x=-13 \end{gathered}[/tex]Finally, the solutions to the equation are:
[tex]\begin{gathered} x_1=1 \\ \\ x_2=-13 \end{gathered}[/tex]