2y(2y-5) + y(2y-5) + (5- 2y)= 0
To solve the equation \( 2y(2y-5) + y(2y-5) + (5 - 2y) = 0 \), we first notice that the first two terms share a common factor \( (2y-5) \). We can factor that out:
\[ (2y - 5)(2y + y) + (5 - 2y) = 0 \]
This simplifies to:
\[ (2y - 5)(3y) + (5 - 2y) = 0 \]
Now, we can expand this a bit to make sure we understand the equation:
\[ 6y^2 - 15y + 5 - 2y = 0 \]
Combining like terms gives us:
\[ 6y^2 - 17y + 5 = 0 \]
Now, we can use the quadratic formula to solve for \( y \). The quadratic formula is given by:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our case, \( a = 6 \), \( b = -17 \), and \( c = 5 \). So we calculate:
- Calculate \( b^2 - 4ac \):
\[ (-17)^2 - 4 \cdot 6 \cdot 5 = 289 - 120 = 169 \]
- Now plug the values into the quadratic formula:
\[ y = \frac{-(-17) \pm \sqrt{169}}{2 \cdot 6} = \frac{17 \pm 13}{12} \]
This gives us two potential solutions:
\[ y = \frac{30}{12} = \frac{5}{2} \quad \text{and} \quad y = \frac{4}{12} = \frac{1}{3} \]
Thus, the solutions to the equation are:
\[ y = \frac{5}{2} \quad \text{or} \quad y = \frac{1}{3} \]
