https://file.aiquickdraw.com/mathbot/file/1726262958352zucmllow.jpg

The image contains two quadratic equations:

1. \( 2x^2 + 9x + 9 = 0 \)
2. \( 2x^2 - 3x - 6x + 9 = 0 \), which simplifies to \( 2x^2 - 9x + 9 = 0 \).

Let's solve both equations.

### First Equation: 
\[
2x^2 + 9x + 9 = 0
\]

We'll use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \(a = 2\), \(b = 9\), and \(c = 9\).

\[
x = \frac{-9 \pm \sqrt{9^2 - 4(2)(9)}}{2(2)}
\]
\[
x = \frac{-9 \pm \sqrt{81 - 72}}{4}
\]
\[
x = \frac{-9 \pm \sqrt{9}}{4}
\]
\[
x = \frac{-9 \pm 3}{4}
\]

So, the two solutions are:
\[
x_1 = \frac{-9 + 3}{4} = \frac{-6}{4} = -\frac{3}{2}
\]
\[
x_2 = \frac{-9 - 3}{4} = \frac{-12}{4} = -3
\]

### Second Equation:
The second equation simplifies to:
\[
2x^2 - 9x + 9 = 0
\]
Using the quadratic formula again, with \(a = 2\), \(b = -9\), and \(c = 9\):

\[
x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(2)(9)}}{2(2)}
\]
\[
x = \frac{9 \pm \sqrt{81 - 72}}{4}
\]
\[
x = \frac{9 \pm \sqrt{9}}{4}
\]
\[
x = \frac{9 \pm 3}{4}
\]

So, the two solutions are:
\[
x_1 = \frac{9 + 3}{4} = \frac{12}{4} = 3
\]
\[
x_2 = \frac{9 - 3}{4} = \frac{6}{4} = \frac{3}{2}
\]

### Summary of Solutions:
1. For the first equation \( 2x^2 + 9x + 9 = 0 \):  
   \(x_1 = -\frac{3}{2}, x_2 = -3\)
2. For the second equation \( 2x^2 - 9x + 9 = 0 \):  
   \(x_1 = 3, x_2 = \frac{3}{2}\)

Would you like any details or further explanation? Here are some related questions:

1. How can you factorize these quadratic equations?
2. Can you verify the solutions graphically?
3. How does the discriminant determine the nature of the solutions?
4. What happens if you substitute the solutions back into the equations?
5. How does the quadratic formula generalize for higher-degree polynomials?

**Tip:** Always simplify equations before applying formulas to make solving easier.

Solve the two quadratic equations: 2x^2 + 9x + 9 = 0 and 2x^2 - 3x - 6x + 9 = 0

Learn how to solve the quadratic equations 2x^2 + 9x + 9 = 0 and 2x^2 - 9x + 9 = 0 using the quadratic formula. The step-by-step solution covers algebraic techniques, quadratic concepts, and solving with the discriminant theorem. Suitable for students in Grades 9-12.

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