To solve the quadratic equation y² + 4y + 4 = 7, we first need to set the equation to zero. We can do this by subtracting 7 from both sides:
y² + 4y + 4 – 7 = 0
This simplifies to:
y² + 4y – 3 = 0
Next, we can use the quadratic formula, which is:
y = (-b ± √(b² – 4ac)) / 2a
In our equation, a = 1, b = 4, and c = -3. Plugging these values into the formula gives:
y = (–4 ± √(4² – 4 × 1 × (-3))) / (2 × 1)
Calculating the discriminant:
4² – 4 × 1 × (-3) = 16 + 12 = 28
Now, substituting this back into the formula:
y = (–4 ± √28) / 2
Since √28 can be simplified to 2√7, we have:
y = (–4 ± 2√7) / 2
This further simplifies to:
y = –2 ± √7
Thus, the two possible values for y are:
y = –2 + √7 and y = –2 – √7
In conclusion, the solutions for the quadratic equation y² + 4y + 4 = 7 are:
- y = -2 + √7
- y = -2 – √7