Question

(2x-3)(3x-2)=(x-2)+(5x+2)(x-4)​

Answer 1

(6x^2-4x)-(9x-6)=x-2+(5x^2-20x)+(2x-8)

6x^2-4x-9x-6=x-2+5x^2-20x+2x-8

6x^2-13x-6=5x^2-17x-10

x^2+14x+4=0

Answer 2

6x²-4x-9x+6=x-2+5x²-20x+2x-8

6x²-13x+6=-17x-10+5x²

6x²-5x²-13x+17x+6+10=0

x²+4x+16=0

D=4²-4×1×16=-48

Нет решений, так как корень из отрицательного числа не существует.

Answer 3

To solve the equation, let's first simplify both sides:

Left-hand side (LHS): (2x - 3)(3x - 2)

Using the distributive property, we can expand this expression: LHS = 2x * 3x + 2x * (-2) - 3 * 3x + (-3) * (-2) LHS = 6x^2 - 4x - 9x + 6 LHS = 6x^2 - 13x + 6

Right-hand side (RHS): (x - 2) + (5x + 2)(x - 4)

Now, let's multiply the terms in the second part of the RHS: RHS = (x - 2) + (5x * x + 5x * (-4) + 2 * x + 2 * (-4)) RHS = (x - 2) + (5x^2 - 20x + 2x - 8) RHS = (x - 2) + (5x^2 - 18x - 8)

Now combine like terms on the RHS: RHS = x - 2 + 5x^2 - 18x - 8 RHS = 5x^2 - 17x - 10

So now the equation becomes: 6x^2 - 13x + 6 = 5x^2 - 17x - 10

Next, let's move all the terms to one side of the equation to set it to zero: 6x^2 - 13x + 6 - 5x^2 + 17x + 10 = 0

Simplify: x^2 + 4x + 16 = 0

At this point, the equation is in quadratic form. To solve for x, we can either factor it or use the quadratic formula. Unfortunately, this quadratic equation does not have real solutions, and it can be confirmed by calculating the discriminant (b^2 - 4ac) which is negative in this case.

Thus, the original equation: (2x - 3)(3x - 2) = (x - 2) + (5x + 2)(x - 4)

Does not have real solutions for x.