Question

(x-6)(x+6)+x(8-2x)+20=0

Answer 1

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(x-6)(x+6)+x(8-2x)+20=0

х^2-36+8х-2х^2+20=0

-х^2-16+8х=0

х^2-8х+16=0

х=-(-8)+-√-8)^2-4*1*16/2*1

х=8+-√64-64/2

х=8+-0/2

х=8/2

х=4

Ответ:х=4

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Answer 2

Ответ:

$- x ^{2} + 8x - 16$

Answer 3

To solve the quadratic equation $(x-6)(x+6) + x(8-2x) + 20 = 0$, you can follow these steps:

1. Expand and simplify the equation: $(x-6)(x+6) + x(8-2x) + 20 = 0$ $x^2 - 36 + x(8-2x) + 20 = 0$

2. Distribute $x$ in the second term: $x^2 - 36 + 8x - 2x^2 + 20 = 0$

3. Combine like terms: $-x^2 + 8x - 16 = 0$

4. Move all terms to one side of the equation to set it equal to zero: $-x^2 + 8x - 16 = 0$

5. To make it easier to work with, multiply the entire equation by -1 to make the leading coefficient positive: $x^2 - 8x + 16 = 0$

6. Now, try to factor this quadratic equation: $(x-4)(x-4) = 0$

7. Since both factors are the same, this equation has a double root at $x = 4$.

So, the solution to the equation $(x-6)(x+6) + x(8-2x) + 20 = 0$ is $x = 4$.