6y² = 17y -12
6 y^2 -17 y = -12
y^2 - (17/6) y = -2
y^2 - (17/6) y + (17/12)^2 = -2 + 289/144
(y- 17/12)^2 = -288/144 + 289/144 = 1/144
y -17/12 = +/- 1/12
y = 18/12 or y = 16/12
To solve the equation 6y² = 17y - 12, we need to rearrange it into a quadratic equation in standard form, which is ax² + bx + c = 0.
Start by moving all the terms to one side of the equation:
6y² - 17y + 12 = 0
Now, we can factorize or use the quadratic formula to find the values of y that satisfy this equation. Let's use the quadratic formula method in this case:
The quadratic formula is given by:
y = (-b ± √(b² - 4ac)) / (2a)
For our equation 6y² - 17y + 12 = 0, we have:
a = 6, b = -17, and c = 12
Plugging these values into the quadratic formula:
y = (-(-17) ± √((-17)² - 4(6)(12))) / (2(6))
Simplifying further:
y = (17 ± √(289 - 288)) / 12
y = (17 ± √1) / 12
Since √1 is equal to 1:
y = (17 ± 1) / 12
This gives us two possible solutions:
1) y = (17 + 1) / 12 = 18 / 12 = 3/2
2) y = (17 - 1) / 12 = 16 / 12 = 4/3
Therefore, the solutions to the equation 6y² = 17y - 12 are y = 3/2 and y = 4/3.