Given the quadratic equation:
![x^2+12x=13](https://img.qammunity.org/2023/formulas/mathematics/college/3hmrsiivnrn6k2pta15973grnocz0r44wl.png)
We can rewrite the equation as follows:
![\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}](https://img.qammunity.org/2023/formulas/mathematics/college/smv4eigknvxdy192rg38vakojjv2n5hqtq.png)
We see that the term inside the parenthesis is a perfect square polynomial. Then:
![(x+6)^2-49=0](https://img.qammunity.org/2023/formulas/mathematics/college/vogiao097nckws6lxsym26fwsujz5e2mv6.png)
Solving for x:
![(x+6)^2=49](https://img.qammunity.org/2023/formulas/mathematics/college/3g8om3v913epj8mq3per3dw5f4os54lqee.png)
Taking the square on both sides:
![\begin{gathered} √((x+6)^2)=√(49) \\ \\ |x+6|=7 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/jp8xzl4cziinsjl69okwwcijdj5j2potnl.png)
This equation can turn into two equations:
![\begin{gathered} x+6=7...(1) \\ x+6=-7...(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/aoao99d8zx9s3u2zvk2gg0xic2ec6np1y6.png)
Solving (1):
![\begin{gathered} x=7-6 \\ \\ \Rightarrow x=1 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/7ypbr30mzy35bfsmdxl6hxfb2z04zd6esg.png)
Solving (2):
![\begin{gathered} x=-7-6 \\ \\ \Rightarrow x=-13 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/d8smv3anapijvfltrxn6fojwfumvmoxmla.png)
Finally, the solutions to the equation are:
![\begin{gathered} x_1=1 \\ \\ x_2=-13 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/jpp6mffo1nki7lg9iwm5pg7ype196jeyje.png)