According to the secant-tangent theorem, we have the following expression:
[tex](x+3)^2=10.8(19.2+10.8)[/tex]
Now, we solve for x.
[tex]\begin{gathered} x^2+6x+9=10.8(30) \\ x^2+6x+9=324 \\ x^2+6x+9-324=0 \\ x^2+6x-315=0 \end{gathered}[/tex]
Then, we use the quadratic formula:
[tex]x_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]
Where a = 1, b = 6, and c = -315.
[tex]\begin{gathered} x_{1,2}=\frac{-6\pm\sqrt[]{6^2-4\cdot1\cdot(-315)}}{2\cdot1} \\ x_{1,2}=\frac{-6\pm\sqrt[]{36+1260}}{2}=\frac{-6\pm\sqrt[]{1296}}{2} \\ x_{1,2}=\frac{-6\pm36}{2} \\ x_1=\frac{-6+36}{2}=\frac{30}{2}=15 \\ x_2=\frac{-6-36}{2}=\frac{-42}{2}=-21 \end{gathered}[/tex]
Hence, the answer is 15 because lengths can't be negative.
