2y^2 3y-6y 4=0?

All_for_one

2 years ago

The equation you've written appears to be:

2y^2 - 3y - 6y + 4 = 0

To solve this equation, you can first combine like terms:

2y^2 - 9y + 4 = 0

Then, you can either use the quadratic formula or factor the equation if possible. If you attempt to factor it, you would look for two numbers that multiply to the product of the leading coefficient (2) and the constant term (4) which is 8, and add up to the middle coefficient (-9). Unfortunately, in this case, there are no such numbers, so you'd likely need to use the quadratic formula:

y = (-b ± √(b² - 4ac)) / (2a)

In your equation, a = 2, b = -9, and c = 4. Plugging these values into the formula will give you the solutions for 'y'.


2y^2 - 9y + 4 = 0

We can use the quadratic formula:

y = (-b ± √(b² - 4ac)) / (2a)

In this equation, a = 2, b = -9, and c = 4. Plugging in these values:

y = (9 ± √((-9)² - 4 * 2 * 4)) / (2 * 2)

y = (9 ± √(81 - 32)) / 4

y = (9 ± √49) / 4

Now, we have two possible solutions:

1. y = (9 + 7) / 4 = 16 / 4 = 4
2. y = (9 - 7) / 4 = 2 / 4 = 0.5

So, the solutions to the equation are y = 4 and y = 0.5.