Question

(3x – 2) - (x + 3) - 3x(x + 1) – 3 = 0;​

Answer 1

(3x – 2) - (x + 3) - 3x(x + 1) – 3 = 0;

3x–2-x-3-3x^2-1–3x=0

3x^2-x-6=0

D=1+6*3=19

нет корней

Answer 2

вот решила как могла.

с помощью дискриминанта

Answer 3

To solve the equation (3x – 2) - (x + 3) - 3x(x + 1) – 3 = 0, we'll follow these steps:

Step 1: Simplify the expression. Step 2: Combine like terms. Step 3: Solve for x.

Let's start:

Step 1: Simplify the expression. (3x - 2) - (x + 3) - 3x(x + 1) - 3

Step 2: Expand the third term (3x(x + 1)): (3x - 2) - (x + 3) - (3x^2 + 3x) - 3

Step 3: Combine like terms: 3x - 2 - x - 3 - 3x^2 - 3x - 3

Now, group the x terms together: (3x - x - 3x) - 2 - 3 - 3x^2 - 3

Simplify further:

• x - 2 - 3 - 3x^2 - 3

Combine constants:

• x - 8 - 3x^2

Step 4: Set the equation equal to zero and arrange it in descending order: -3x^2 - x - 8 = 0

Step 5: Now, to solve the quadratic equation, we can use the quadratic formula: The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:

x = (-b ± √(b^2 - 4ac)) / 2a

In our case, a = -3, b = -1, and c = -8.

Substitute the values into the formula: x = (1 ± √((-1)^2 - 4 * (-3) * (-8))) / 2 * (-3)

Simplify further: x = (1 ± √(1 - 96)) / (-6)

x = (1 ± √(-95)) / (-6)

Since the term inside the square root is negative, there are no real number solutions for x. The equation has complex solutions, but we won't be able to find them using the real number system. Hence, the equation (3x – 2) - (x + 3) - 3x(x + 1) – 3 = 0 has no real number solutions.