Question

(2x+5)(x+2)=21 решите пж

Answer 1

Фоточка тебе в помощь ;)

Answer 2

To solve the equation (2x + 5)(x + 2) = 21, you need to find the values of x that satisfy the equation. We'll start by expanding the left-hand side (LHS) of the equation:

(2x + 5)(x + 2) = 21 2x^2 + 4x + 5x + 10 = 21 2x^2 + 9x + 10 = 21

Now, we'll move all terms to one side to set the equation equal to zero:

2x^2 + 9x + 10 - 21 = 0 2x^2 + 9x - 11 = 0

Now, to solve this quadratic equation, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

Where a, b, and c are the coefficients of the quadratic equation (ax^2 + bx + c = 0).

In our equation, a = 2, b = 9, and c = -11.

x = (-(9) ± √(9^2 - 4 * 2 * -11)) / 2 * 2 x = (-9 ± √(81 + 88)) / 4 x = (-9 ± √169) / 4 x = (-9 ± 13) / 4

Now, we have two possible solutions for x:

1. x = (-9 + 13) / 4 x = 4 / 4 x = 1

2. x = (-9 - 13) / 4 x = -22 / 4 x = -5.5

So, the solutions to the equation (2x + 5)(x + 2) = 21 are x = 1 and x = -5.5.