Question

4y²−28y+49=0 ;y = ?​

Answer 1

Ответ: .)(2y-7)²=0
2y-7=0
2y=7
y=3,5

Объяснение:

надеюсь помогла

Answer 2

To solve the quadratic equation 4y² - 28y + 49 = 0, you can use the quadratic formula:

y = (-b ± √(b² - 4ac)) / (2a)

In this equation, a = 4, b = -28, and c = 49. Plug these values into the formula:

y = (-(-28) ± √((-28)² - 4 * 4 * 49)) / (2 * 4)

Now, calculate the values inside the square root:

y = (28 ± √(784 - 784)) / 8

Since the discriminant (the value inside the square root) is zero, there is only one real root for this equation:

y = (28 ± √0) / 8

y = (28 ± 0) / 8

y = 28 / 8

Simplify the fraction:

y = 7/2

So, the solution to the equation 4y² - 28y + 49 = 0 is:

y = 7/2