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Question 1184193: 9x^2+72x-128y+80=0
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! To solve for y in the equation 9x² + 72x - 128y + 80 = 0, we need to isolate y on one side of the equation. Here's how:
1. **Move the terms without y to the other side:**
Subtract 9x² and 72x and 80 from both sides of the equation:
-128y = -9x² - 72x - 80
2. **Divide by -128 to solve for y:**
y = (-9x² - 72x - 80) / -128
3. **Simplify (optional):**
You can simplify the equation by dividing each term in the numerator by -128:
y = (9/128)x² + (72/128)x + (80/128)
y = (9/128)x² + (9/16)x + (5/8)
Therefore, the solution for y is:
y = (9/128)x² + (9/16)x + (5/8)
or
y = (9x² + 72x + 80) / 128
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