Question

 3*(х+1)2=2х+2  0,1х2-3х-5=0.

Answer 1

Ответ:-1

Пошаговое объяснение:

3*(х+1)2=2х+2

6x+6= 2x+2

6x-2x=2-6

4x=-4

x=-1

надеюсь понятно

Answer 2

To solve the equations you've provided, let's break it down step by step:

1. Solve the first equation: 3*(x+1)^2 = 2x + 2

First, expand the left side of the equation:

3(x^2 + 2x + 1) = 2x + 2

Now, distribute the 3 on the left side:

3x^2 + 6x + 3 = 2x + 2

Next, move all terms to one side to set the equation equal to zero:

3x^2 + 6x + 3 - 2x - 2 = 0

Simplify:

3x^2 + 4x + 1 = 0

2. Solve the second equation: 0.1x^2 - 3x - 5 = 0

To make it easier to work with, let's first multiply the entire equation by 10 to get rid of the decimal:

10(0.1x^2 - 3x - 5) = 10(0)

x^2 - 30x - 50 = 0

Now you have a quadratic equation:

x^2 - 30x - 50 = 0

To solve both equations, we can use the quadratic formula:

For a quadratic equation of the form ax^2 + bx + c = 0, the solutions for x are given by:

x = (-b ± √(b² - 4ac)) / (2a)

1. For the first equation (3x^2 + 4x + 1 = 0):

a = 3, b = 4, and c = 1

Using the quadratic formula:

x = (-4 ± √(4² - 4 * 3 * 1)) / (2 * 3) x = (-4 ± √(16 - 12)) / 6 x = (-4 ± √4) / 6 x = (-4 ± 2) / 6

Now, calculate the two possible solutions:

x₁ = (-4 + 2) / 6 x₁ = -2 / 6 x₁ = -1/3

x₂ = (-4 - 2) / 6 x₂ = -6 / 6 x₂ = -1

So, the solutions for the first equation are x = -1/3 and x = -1.

2. For the second equation (x^2 - 30x - 50 = 0):

a = 1, b = -30, and c = -50

Using the quadratic formula:

x = (30 ± √((-30)² - 4 * 1 * (-50))) / (2 * 1) x = (30 ± √(900 + 200)) / 2 x = (30 ± √1100) / 2

Now, calculate the two possible solutions:

x₁ = (30 + √1100) / 2 x₂ = (30 - √1100) / 2

These are the solutions for the second equation. You can use a calculator to approximate the values for x₁ and x₂.