Question

(x^2-3x)^2-8(x^2-3x)-20=0

Answer 1

Ответ:мне лень было на клавиатуре все это писать,вот тебе решение на листочке.ответ 5,-2,2,1

Объяснение:

Answer 2

Let's simplify the equation first by setting y = x^2 - 3x. This substitution makes the equation look like:

y^2 - 8y - 20 = 0

Now, we can solve for y using the quadratic formula:

y = (8 ± √(8^2 - 4(1)(-20))) / 2(1) y = (8 ± √144) / 2 y = 4 ± 6

So, y = 10 or y = -2.

Substituting back in for y gives us:

x^2 - 3x = 10 or x^2 - 3x = -2

We can solve for x using the quadratic formula for each equation:

For x^2 - 3x = 10: x = (3 ± √49) / 2 x = (3 ± 7) / 2 x = 5 or x = -2

For x^2 - 3x = -2: x = (3 ± √13) / 2

Therefore, the solutions to the original equation are x = 5, x = -2, x = (3 + √13) / 2, and x = (3 - √13) / 2.