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x^2-20x+99=0  x = 11 and x = 9

$\begin{array}{rcl} x^2-20x+99 &=&0 \\ \boxed{~ x = {-b \pm \sqrt{b^2-4ac} \over 2a} ~} \\ x &=& {20 \pm \sqrt{20^2-4\cdot 1 \cdot 99 } \over 2} \qquad | \qquad a = 1 \quad b = -20 \quad c = 99 \\ x &=& {20 \pm \sqrt{400-396 } \over 2} \\ x &=& {20 \pm \sqrt{4 } \over 2} \\ x &=& {20 \pm 2 \over 2} \\ x_1 &=& {20 + 2 \over 2} \\ x_1 &=& {22 \over 2} \\ \mathbf{x_1} &\mathbf{=}& \mathbf{11} \\\\ x_2 &=& {20 - 2 \over 2} \\ \mathbf{x_2} &\mathbf{=}& \mathbf{9}\\ \end{array}$

Hallo guest

This can be factored as (x - 9)(x - 11) = 0 so the solutions are x = 9 and x = 11