To solve the equation 6y² + 2y + 1 = 3y² + 6, we start by rearranging it to one side of the equation:
1. Move all terms to one side:
6y² + 2y + 1 – 3y² – 6 = 0
2. Combine like terms:
(6y² – 3y²) + 2y + (1 – 6) = 0
This simplifies to:
3y² + 2y – 5 = 0
3. Now, we can apply the quadratic formula, which is:
y = (-b ± √(b² – 4ac)) / 2a
In our equation, a = 3, b = 2, and c = -5.
4. First, calculate the discriminant (b² – 4ac):
Discriminant = 2² – 4(3)(-5) = 4 + 60 = 64
5. Now substitute back into the quadratic formula:
y = (-2 ± √64) / (2 * 3)
This simplifies to:
y = (-2 ± 8) / 6
6. Now find the two potential solutions:
For y = (-2 + 8) / 6 = 6 / 6 = 1
And for y = (-2 – 8) / 6 = -10 / 6 = -5/3
7. Thus, the solutions to the equation are:
y = 1 and y = -5/3.