y²=4y+96
To solve the equation \( y^2 = 4y + 96 \), we can first rearrange it to the standard form of a quadratic equation. This involves moving all terms to one side of the equation:
\[ y^2 - 4y - 96 = 0 \]
Now, we can solve this quadratic equation using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our case, \( a = 1 \), \( b = -4 \), and \( c = -96 \).
- First, we calculate the discriminant \( b^2 - 4ac \):
\[ b^2 = (-4)^2 = 16 \] \[ 4ac = 4 \cdot 1 \cdot (-96) = -384 \] \[ b^2 - 4ac = 16 - (-384) = 16 + 384 = 400 \]
- Now we substitute the values into the quadratic formula:
\[ y = \frac{-(-4) \pm \sqrt{400}}{2 \cdot 1} \] \[ y = \frac{4 \pm 20}{2} \]
- This gives us two possible solutions:
\[ y_1 = \frac{4 + 20}{2} = \frac{24}{2} = 12 \] \[ y_2 = \frac{4 - 20}{2} = \frac{-16}{2} = -8 \]
Thus, the solutions to the quadratic equation \( y^2 = 4y + 96 \) are:
\[ y = 12 \quad \text{and} \quad y = -8 \]
