Question

(2x-3)(3x-4)-(3x+1)(x-2)-12x+31=0

Answer 1

$(2x-3)(3x-4)-(3x+1)(x-2)-12x+31=0\\\\\\6x^2-8x-9x+12-(3x^2-6x+x-2)-12x+31=0\\\\\\6x^2-8x-9x+12-3x^2+5x+2-12x+31=0\\\\\\3x^2-24x+45=0\;\;\;|\div3\\\\\\x^2-8x+15=0\\\\\\D=(-8)^2-4\cdot1\cdot15=64-60=4\\\\\\x_{1}=\dfrac{8-2}{2}=\dfrac{6}{2}=3\\\\\\x_2=\dfrac{8+2}{2}=\dfrac{10}{2}=5\\\\\\Otvet:\;x_1=3;\;x_2=5.$

Answer 2

To solve the given equation, you can follow these steps:

1. Expand the expressions within the parentheses.
2. Combine like terms.
3. Isolate the variable (in this case, 'x') on one side of the equation.

Let's go through these steps:

(2x - 3)(3x - 4) - (3x + 1)(x - 2) - 12x + 31 = 0

1. Expand the expressions within the parentheses:

(2x * 3x) + (2x * -4) + (-3 * 3x) + (-3 * -4) - (3x * x) - (3x * -2) - (1 * x) - (1 * -2) - 12x + 31 = 0

This simplifies to:

6x^2 - 8x - 9x + 12 - 3x^2 + 6x - x + 2 - 12x + 31 = 0

2. Combine like terms:

(6x^2 - 3x^2) + (-8x - 9x - x - 12x + 6x) + (12 + 2 + 31) = 0

3x^2 - 22x + 45 = 0

3. Now, you have a quadratic equation in the form of ax^2 + bx + c = 0, where a = 3, b = -22, and c = 45. To solve for 'x,' you can use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

x = (-(-22) ± √((-22)² - 4 * 3 * 45)) / (2 * 3)

x = (22 ± √(484 - 540)) / 6

x = (22 ± √(-56)) / 6

Since the discriminant (the value inside the square root) is negative, the equation has complex solutions. You can simplify it further:

x = (22 ± √(56i)) / 6

Now, you have two complex solutions for 'x':

x = (22 + √(56i)) / 6 x = (22 - √(56i)) / 6

These are the solutions to the given equation.