Question

4(3y+1)^2-27 рівняння

Answer 1

________________

Готово!!Удачи))

_________________

Answer 2

4(3у+1)^2-27=

4(9у^2+6у+1)-27=

36у^2+24у+4-27=

36у^2+24у-23

Answer 3

To solve the equation 4(3y+1)^2 - 27 = 0, follow these steps:

Step 1: Expand the square term. 4(3y+1)^2 - 27 = 0 4(9y^2 + 6y + 1) - 27 = 0

Step 2: Distribute the 4 to each term inside the parentheses. 36y^2 + 24y + 4 - 27 = 0

Step 3: Combine like terms. 36y^2 + 24y - 23 = 0

Now, you have a quadratic equation in the form of ay^2 + by + c = 0, where a = 36, b = 24, and c = -23.

Step 4: Solve the quadratic equation. There are a couple of methods to do this. You can use the quadratic formula or factor the equation if it can be factored easily. Let's use the quadratic formula:

The quadratic formula is given by: y = (-b ± √(b^2 - 4ac)) / 2a

In our equation, a = 36, b = 24, and c = -23.

y = (-(24) ± √(24^2 - 4 * 36 * -23)) / 2 * 36

y = (-24 ± √(576 + 3312)) / 72

y = (-24 ± √3888) / 72

y = (-24 ± 62.33) / 72

Now, we have two solutions:

1. y = (-24 + 62.33) / 72 y = 38.33 / 72 y ≈ 0.5324

2. y = (-24 - 62.33) / 72 y = -86.33 / 72 y ≈ -1.1986

So, the solutions to the equation are approximately: y ≈ 0.5324 and y ≈ -1.1986