Question

3х(х-2)-х(3+х)=-2(х-4)+12​

Answer 1

Ответ:

Думаю понятно, формула сбоку))

Answer 2

To solve the given equation $3x(x-2) - x(3+x) = -2(x-4) + 12$, we'll follow these steps:

Step 1: Distribute and simplify on both sides of the equation. Step 2: Combine like terms. Step 3: Solve for $x$.

Let's proceed with these steps:

Step 1: $3x(x-2) - x(3+x) = -2(x-4) + 12$

Expand the terms on both sides: $3x^2 - 6x - 3x^2 - x^2 = -2x + 8 + 12$

Step 2: Combine like terms: $-4x^2 - 6x = -2x + 20$

Step 3: Move all terms to one side to set the equation equal to zero: $-4x^2 - 6x + 2x - 20 = 0$

Simplify further: $-4x^2 - 4x - 20 = 0$

Now, divide the entire equation by -4 to simplify and change the sign of all terms: $x^2 + x + 5 = 0$

Now, we have a quadratic equation in the form $ax^2 + bx + c = 0$, where $a = 1$, $b = 1$, and $c = 5$. We can use the quadratic formula to solve for $x$:

The quadratic formula is: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substitute the values of $a$, $b$, and $c$: $x = \frac{-1 \pm \sqrt{1 - 4 \times 1 \times 5}}{2 \times 1}$

Simplify under the square root: $x = \frac{-1 \pm \sqrt{-19}}{2}$

Since the discriminant ($-19$) is negative, the solutions will be complex numbers. Let's express them in terms of $i$ (the imaginary unit):

$x = \frac{-1 \pm \sqrt{19i}}{2}$

So, the solutions for $x$ in terms of complex numbers are: $x = \frac{-1 + \sqrt{19i}}{2}$ and $x = \frac{-1 - \sqrt{19i}}{2}$