Respuesta :
[tex]ax^2+bx+c=0[/tex]
To complete the square you add and subtract the next to the equation:
[tex](\frac{b}{2})^2[/tex]For the given equation:
[tex]x^2+12x+2=0[/tex][tex](\frac{b}{2})^2=(\frac{12}{2})^2=6^2[/tex]Add and subtract 6 squared as follow:
[tex]x^2+12x+6^2+2-6^2=0[/tex]Simplify:
[tex]\begin{gathered} (a^2+2ab+b^2)=(a+b)^2 \\ \\ x^2+12x+6^2=(x+6)^2 \\ \\ \\ \\ (x+6)^2+2-6^2=0 \\ (x+6)^2+2-36=0 \\ (x+6)^2-34=0 \end{gathered}[/tex]Solve x by taking square roots:
[tex]\begin{gathered} (x+6)^2-34+34=0+34 \\ (x+6)^2=34 \\ \\ \sqrt[]{(x+6)^2}=\sqrt[]{34} \\ \\ x+6=\pm\sqrt[]{34} \\ x+6-6=\pm\sqrt[]{34}-6 \\ \\ x_1=\sqrt[]{34}-6 \\ x_2=-\sqrt[]{34}-6 \end{gathered}[/tex]Then, the solutions for the given quadratic equation are:[tex]\begin{gathered} x_1=\sqrt[]{34}-6 \\ x_2=-\sqrt[]{34}-6 \end{gathered}[/tex]