Solve 6y2=17y-12
6y^2 - 17y + 12 = 0
(3y-4)(2y-3) = 0
now finish it off
6y^2 - 17y + 12 = 0
(3y - 4)(2y - 3) = 0
y = 4/3 or y = 3/2
I assume you would know how to factor this trinomial
To solve the equation 6y^2 = 17y - 12, we need to rearrange it in the form of a quadratic equation, which is ax^2 + bx + c = 0. In this case, we already have the equation in that form.
So, the first step is to set it equal to zero by subtracting 17y and adding 12 to both sides:
6y^2 - 17y + 12 = 0
Now, we can factor or use the quadratic formula to solve for y.
To factor the quadratic equation, we need to find two numbers that multiply to give us 6 * 12 = 72 and add up to -17 (the coefficient of the middle term).
The two numbers that satisfy these conditions are -8 and -9. Therefore, we can factor the quadratic equation:
(2y - 3)(3y - 4) = 0
Now, we can set each factor equal to zero and solve for y:
2y - 3 = 0 OR 3y - 4 = 0
When we solve the first equation, we get:
2y = 3
y = 3/2
And when we solve the second equation, we get:
3y = 4
y = 4/3
So, the solutions to the equation 6y^2 = 17y - 12 are y = 3/2 and y = 4/3.