Question 12 Solve for x . If there is more than one solution, separate the solutions with a comma. \( \frac{8}{x-4}+\frac{x}{x+5}=\frac{-6 x+16}{x^{2}+x-20} \) \( x=\square \)

To solve the equation \( \frac{8}{x-4}+\frac{x}{x+5}=\frac{-6 x+16}{x^{2}+x-20} \), first notice that the denominator on the right side can be factored as \( (x-4)(x+5) \). Clearing the denominators, we multiply both sides by \( (x-4)(x+5) \) to get rid of the fractions: \[ 8(x+5) + x(x-4) = -6x + 16 \] Now, distribute and simplify: \[ 8x + 40 + x^2 - 4x = -6x + 16 \] Combine like terms: \[ x^2 + 10x + 40 = -6x + 16 \] \[ x^2 + 16x + 24 = 0 \] Next, solve this quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Where \( a = 1, b = 16, c = 24 \): \[ x = \frac{-16 \pm \sqrt{16^2 - 4 \cdot 1 \cdot 24}}{2 \cdot 1} \] \[ x = \frac{-16 \pm \sqrt{256 - 96}}{2} \] \[ x = \frac{-16 \pm \sqrt{160}}{2} \] \[ x = \frac{-16 \pm 4\sqrt{10}}{2} \] \[ x = -8 \pm 2\sqrt{10} \] Thus, the solutions are: \( x = -8 + 2\sqrt{10}, -8 - 2\sqrt{10} \) So, \( x=\square \) is \( -8 + 2\sqrt{10}, -8 - 2\sqrt{10} \).