To solve the quadratic equation \(-9x^2 - 2x + 3 = 0\), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = -9\), \(b = -2\), and \(c = 3\).
First, we calculate the discriminant (\(b^2 - 4ac\)):
\[ b^2 - 4ac = (-2)^2 - 4(-9)(3) = 4 + 108 = 112 \]
Since the discriminant is positive, there are two real solutions. Let's now compute the two solutions:
\[ x = \frac{-(-2) \pm \sqrt{112}}{2 \cdot (-9)} = \frac{2 \pm \sqrt{112}}{-18} \]
Next, simplify \(\sqrt{112}\):
\[ \sqrt{112} = \sqrt{16 \cdot 7} = 4\sqrt{7} \]
Now substitute this back into the equation:
\[ x = \frac{2 \pm 4\sqrt{7}}{-18} \]
Which can be simplified by dividing the numerator and denominator by 2:
\[ x = \frac{1 \pm 2\sqrt{7}}{-9} \]
So the two solutions are:
\[ x_1 = \frac{1 + 2\sqrt{7}}{-9} \quad \text{and} \quad x_2 = \frac{1 - 2\sqrt{7}}{-9} \]
Using the negative sign, we can rewrite these as:
\[ x_1 = -\frac{1 + 2\sqrt{7}}{9} \quad \text{and} \quad x_2 = -\frac{1 - 2\sqrt{7}}{9} \]
Thus, the exact solutions are:
- Solution 1 (Box 1): \(-\frac{1 + 2\sqrt{7}}{9}\)
- Solution 2 (Box 2): \(-\frac{1 - 2\sqrt{7}}{9}\)
Here are the solutions formatted as requested:
Box 1: \(-\frac{1 + 2\sqrt{7}}{9}\)
Box 2: \(-\frac{1 - 2\sqrt{7}}{9}\)
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