Solve Complex Equation: (x+8)(8-x) + 4(x-3)(x+3) + 5(6-x²) = 0

To solve the equation $(x+8)(8-x)+4(x-3)(x+3)+5(6-x^2)=0$, we will follow these steps:

• Step 1: Expand and simplify each factor using important algebraic formulas.
• Step 2: Combine all terms to form a quadratic equation.
• Step 3: Solve the quadratic equation using the quadratic formula.

Let's work through each step:

Step 1: Expand each part of the equation:

• The first term $(x+8)(8-x)$ is a difference of squares, which simplifies to:
  $(x+8)(8-x) = (8^2 - x^2) = 64 - x^2$.
• The second term $4(x-3)(x+3)$ is another difference of squares:
  $4[(x^2 - 9)] = 4x^2 - 36$.
• The third term $5(6-x^2)$ simplifies to:
  $30 - 5x^2$.

Step 2: Combine the results to form a quadratic equation:

Combine terms in the equation:

$64 - x^2 + 4x^2 - 36 + 30 - 5x^2 = 0$

Simplify further:

$(4x^2 - x^2 - 5x^2) + (64 - 36 + 30) = 0$

$-2x^2 + 58 = 0$

Rearrange to standard quadratic form:

$2x^2 = 58$

Step 3: Solve using the quadratic formula:

The equation simplifies to $x^2 = 29$.

Taking the square root of both sides gives the solutions:

$x = \pm \sqrt{29}$.

Thus, the solution to the equation is $\pm \sqrt{29}$.