この問題教えてください！！

$(3x+13)^2−4(3x+13)−221=0$

$9x^2+78x+169−4(3x+13)−221=0$

$9^2+78x+169−12x−52−221=0$

$9^2+66x+169−52−221=0$

$9^2+66x+117−221=0$

$9x^2+66x−104=0$

$9x^2+66x=104$

$\dfrac{9x^2+66x}{9}=\dfrac{104}{9}$

$x^2+\dfrac{66}{9}x=\dfrac{104}{9}$

$x^2+\dfrac{22}{3}x=\dfrac{104}{9}$

$x^2+\dfrac{22}{3}x+\left(\dfrac{11}{3}\right)^2=\dfrac{104}{9}+\left(\dfrac{11}{3}\right)^2$

$x^2+\dfrac{22}{3}+\dfrac{121}{9}=\dfrac{104+121}{9}$

$x^2+\dfrac{22}{3}x+\dfrac{121}{9}=25$

$\left(x+\dfrac{11}{3}\right)^2=25$

$\sqrt{\left(x+\dfrac{11}{3}^2\right)}=\sqrt{25}$

$x+\dfrac{11}{3}=5$

$x+\dfrac{11}{3}=−5$

〈解〉

$x=\dfrac{4}{3}$

$x=−\dfrac{26}{3}$