Someone teach me step by step

9l9-8xl=2x+3 it saids extraneous solution? can someone teach me step by step to solve this?

$\begin{array}{rcll} 9 |9-8x| &=& 2x+3 \qquad & | \qquad (\text{square both sides}) \\ 9^2 \cdot (9-8x)^2 &=& (2x+3)^2 \\ 81 \cdot (9^2-2\cdot 9 \cdot 8\cdot x +8^2x^2 ) &=& 2^2x^2 +2\cdot 2 \cdot 3\cdot x + 3^2 \\ 81 \cdot (81-144 x +64x^2 ) &=& 4x^2 + 12x + 9 \\ 81^2 - 81\cdot 144 x + 81\cdot 64x^2 &=& 4x^2 + 12x + 9 \\ 6561 - 11664x + 5184x^2 &=& 4x^2 + 12x + 9 \qquad & | \qquad -4x^2\\ 6561 - 11664x + 5180x^2 &=& 12x + 9 \qquad & | \qquad -12x\\ 6561 - 11676x + 5180x^2 &=& 9 \qquad & | \qquad -9\\ 6552 - 11676x + 5180x^2 &=& 0 \\ 5180x^2 - 11676x + 6552 &=& 0 \qquad & | \qquad :4\\ 1295x^2 - 2919x + 1638 &=& 0\\ \end{array}$

$\boxed{~ \begin{array}{rcll} ax^2+bx+c &=& 0 \\ x &=& \dfrac{-b \pm \sqrt{b^2-4ac} }{2a} \end{array} ~}$

$\begin{array}{rcll} 1295x^2 - 2919x + 1638 &=& 0 \quad | \quad a=1295 \quad b = -2919 \quad c = 1638 \\ x &=& \dfrac{-(-2919) \pm \sqrt{(-2919)^2-4\cdot 1295 \cdot 1638} }{2\cdot 1295} \\ x &=& \dfrac{2919 \pm \sqrt{ 8520561-8484840 } }{2590} \\ x &=& \dfrac{2919 \pm \sqrt{ 35721 } }{2590} \\ x &=& \dfrac{2919 \pm 189 }{2590} \\\\ x_1 &=& \dfrac{2919 + 189 }{2590} \\ x_1 &=& \dfrac{3108 }{2590} \\ \mathbf{x_1} & \mathbf{=} & \mathbf{1.2} \\\\ x_2 &=& \dfrac{2919 - 189 }{2590} \\ x_2 &=& \dfrac{2730}{2590} \\ x_2 &=& \dfrac{273}{259} \\ \mathbf{x_2} &\mathbf{=}& \mathbf{1.\overline{054}}\\ \end{array}$