Solve by factorisation . `2x^2+9x+9=0`

Answer

To solve the quadratic equation \(2x^2 + 9x + 9 = 0\) by factorization, we will follow these steps:

### Step 1: Identify coefficients
The equation is in the standard form \(Ax^2 + Bx + C = 0\), where:
- \(A = 2\)
- \(B = 9\)
- \(C = 9\)

### Step 2: Calculate the product of \(A\) and \(C\)
We need to find the product \(A \times C\):
\[ A \times C = 2 \times 9 = 18 \]

### Step 3: Find two numbers that multiply to \(A \times C\) and add to \(B\)
We need to find two numbers that multiply to \(18\) and add up to \(9\). The numbers are \(6\) and \(3\) because:
\[ 6 \times 3 = 18 \quad \text{and} \quad 6 + 3 = 9 \]

### Step 4: Rewrite the middle term using the two numbers
We can rewrite \(9x\) as \(6x + 3x\):
\[ 2x^2 + 6x + 3x + 9 = 0 \]

### Step 5: Group the terms
Now, we group the terms:
\[ (2x^2 + 6x) + (3x + 9) = 0 \]

### Step 6: Factor by grouping
Factor out the common factors in each group:
\[ 2x(x + 3) + 3(x + 3) = 0 \]

### Step 7: Factor out the common binomial
Now we can factor out the common binomial \((x + 3)\):
\[ (2x + 3)(x + 3) = 0 \]

### Step 8: Set each factor to zero
Now we set each factor to zero:
1. \(2x + 3 = 0\)
2. \(x + 3 = 0\)

### Step 9: Solve for \(x\)
For the first equation:
\[ 2x + 3 = 0 \implies 2x = -3 \implies x = -\frac{3}{2} \]

For the second equation:
\[ x + 3 = 0 \implies x = -3 \]

### Final Solution
The solutions to the equation \(2x^2 + 9x + 9 = 0\) are:
\[ x = -3 \quad \text{and} \quad x = -\frac{3}{2} \]

---