Question

(x-1)(x-2)+(x+3)(x-3)+3x+18=0

Answer 1

X^2-2x-x+2+x^2+3x-3x-9+3x+18=0
2x^2+11=0
D=0-4*2*11=0-88=-88
D<0 - действительных корней нет

Answer 2

Let's simplify the given equation:

Expanding the terms: (x - 1)(x - 2) + (x + 3)(x - 3) + 3x + 18 = 0

(x^2 - 3x + 2) + (x^2 - 9) + 3x + 18 = 0

x^2 - 3x + 2 + x^2 - 9 + 3x + 18 = 0

Combining like terms: 2x^2 - 3x^2 - 3x + 3x + 2 - 9 + 18 = 0

-x^2 + 11 = 0

To find the solutions, we can set the equation equal to zero:

-x^2 + 11 = 0

Multiplying through by -1 to make the leading coefficient positive:

x^2 - 11 = 0

Now, we can solve this quadratic equation using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 1, b = 0, and c = -11. Substituting these values into the quadratic formula:

x = (0 ± √(0^2 - 4(1)(-11))) / (2(1))

Simplifying further:

x = (± √(0 + 44)) / 2

x = (± √44) / 2

x = ± √(44) / 2

x = ± (2√11) / 2

Simplifying the expression:

x = ± √11

Therefore, the solutions to the equation (x - 1)(x - 2) + (x + 3)(x - 3) + 3x + 18 = 0 are x = √11 and x = -√11.