Question

2) (y-2) (2 y - 1) = 5.​

Answer 1

Y1=-1/2,y2=3 не знаю навіть,але може це допоможе вам

Answer 2

To solve the equation (y - 2)(2y - 1) = 5, you can follow these steps:

1. Expand the left side of the equation by applying the distributive property (FOIL method) to the terms inside the parentheses:

(y - 2)(2y - 1) = 2y(y) - 2(y) - 1(2y) + 2(1) = 2y^2 - 2y - 2y + 2 = 2y^2 - 4y + 2

So, the equation becomes:

2y^2 - 4y + 2 = 5

2. Now, subtract 5 from both sides of the equation to isolate the terms on the left side:

2y^2 - 4y + 2 - 5 = 0

2y^2 - 4y - 3 = 0

3. Next, you can attempt to factor the quadratic equation. In this case, the equation doesn't easily factor into simple linear terms, so you can use the quadratic formula to solve for y:

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

In the equation 2y^2 - 4y - 3 = 0:

a = 2, b = -4, and c = -3.

Plug these values into the quadratic formula:

y = (-(-4) ± √((-4)^2 - 4(2)(-3))) / (2(2))

y = (4 ± √(16 + 24)) / 4

y = (4 ± √40) / 4

y = (4 ± 2√10) / 4

4. Now, simplify the expression by factoring out a common factor of 2 from both the numerator and the denominator:

y = (2(2 ± √10)) / (2(2))

y = (2 ± √10) / 2

5. Finally, simplify further by dividing both the numerator and denominator by 2:

y = (2 ± √10) / 2 = (1 ± √10) / 1 = 1 ± √10

So, the solutions to the equation (y - 2)(2y - 1) = 5 are:

y = 1 + √10 and y = 1 - √10