in the diagram segment AD and AB are tangent to circle C solve for x

A property ostates that if two lines that are tangent to the circle intersect in an external point, they are congruent, i.e. they have the same length.
[tex]\begin{gathered} AD=AB \\ x^2+2=11 \end{gathered}[/tex]From this expression we can determine the possible values of x. The first step is to equal the expression to zero
[tex]\begin{gathered} x^2+2-11=11-11 \\ x^2+2-11=0 \\ x^2-9 \end{gathered}[/tex]The expression obtained is a quadratic equation, using the queadratic formula we can determine the possible values of x:
[tex]\begin{gathered} f(x)=ax^2+bx+c \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}[/tex]For our expression
[tex]x^2+0x+-9[/tex]The coefficients are
a=1
b=0
c=-9
Replace them in the formula
[tex]\begin{gathered} x=\frac{-0\pm\sqrt[]{0^2-4\cdot1\cdot(-9)}}{2\cdot1} \\ x=\frac{0\pm\sqrt[]{36}}{2} \\ x=\frac{0\pm6}{2} \end{gathered}[/tex]Now calculate both possible values:
Positive:
[tex]\begin{gathered} x=\frac{+6}{2} \\ x=3 \end{gathered}[/tex]Negative:
[tex]\begin{gathered} x=\frac{-6}{2} \\ x=-3 \end{gathered}[/tex]The possible values of x are 3 and -3