We have the next quadratic equation
[tex]x^2-6x-7=0[/tex]in order to have a trinomial perfect square that is
[tex]a^2+2ab+b^2[/tex]in this case, we have
a= x
b= -3
so the expected value
[tex]b^2=(-3)^2=9[/tex]In order to complete the square, we need to add the number 9 on both sides of the equation
[tex]\begin{gathered} x^2-6x-7+9=9 \\ x^2-6x+9=9+7 \\ x^2-6x+9=16 \end{gathered}[/tex][tex](x-3)^2=16[/tex]Because we have a quadratic equation we will have 2 solutions for x
[tex]\begin{gathered} (x-3)=\pm\sqrt[]{16} \\ x-3=\pm4 \\ x_1=4+3=7 \\ x_2=-4+3=-1 \end{gathered}[/tex]the solution is
x1=7
x2=-1
