Question

(x-3)(x+2)->15x+10

Answer 1

Вроде бы так)

Мне кажется правильно)

Answer 2

Let's solve the given equation:

(x - 3)(x + 2) = 15x + 10

To solve this equation, we first need to expand the left-hand side (LHS) using the distributive property:

(x - 3)(x + 2) = x(x) + x(2) - 3(x) - 3(2) = x^2 + 2x - 3x - 6 = x^2 - x - 6

Now we have the equation:

x^2 - x - 6 = 15x + 10

To solve for x, we'll move all the terms to one side of the equation to set it to zero:

x^2 - x - 15x - 6 - 10 = 0

Combine like terms:

x^2 - 16x - 16 = 0

Now, we have a quadratic equation in the standard form (ax^2 + bx + c = 0) where a = 1, b = -16, and c = -16. To solve this quadratic equation, we can use factoring or the quadratic formula.

Using the quadratic formula, which 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

Plugging in the values:

x = (16 ± √((-16)^2 - 4 * 1 * -16)) / 2 * 1 x = (16 ± √(256 + 64)) / 2 x = (16 ± √320) / 2 x = (16 ± √(16 * 20)) / 2 x = (16 ± 4√5) / 2 x = 8 ± 2√5

So, the two solutions for x are:

1. x = 8 + 2√5
2. x = 8 - 2√5

Remember that both of these values are approximate since √5 is an irrational number.