To solve the equation \( 5x^2 - 6x - 11 = x^2 - 9 \), we first simplify both sides.
- Rearranging the equation gives:
\[ 5x^2 - 6x - 11 - x^2 + 9 = 0. \]
- Combine like terms:
\[ (5x^2 - x^2) - 6x + (9 - 11) = 0, \] \[ 4x^2 - 6x - 2 = 0. \]
- Next, we can simplify the equation by dividing all terms by 2:
\[ 2x^2 - 3x - 1 = 0. \]
Now we can use the quadratic formula to solve for \( x \), where \( a = 2 \), \( b = -3 \), and \( c = -1 \):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \]
Substituting in our values:
\[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2}, \] \[ x = \frac{3 \pm \sqrt{9 + 8}}{4}, \] \[ x = \frac{3 \pm \sqrt{17}}{4}. \]
Now we can find the two potential solutions:
- Using the plus sign:
\[ x_1 = \frac{3 + \sqrt{17}}{4} \approx \frac{3 + 4.1231}{4} \approx \frac{7.1231}{4} \approx 1.7808. \]
- Using the minus sign:
\[ x_2 = \frac{3 - \sqrt{17}}{4} \approx \frac{3 - 4.1231}{4} \approx \frac{-1.1231}{4} \approx -0.2808. \]
Now rounding to the nearest tenth:
- For \( x_1 \approx 1.8 \).
- For \( x_2 \approx -0.3 \).
Thus, the solutions to the equation \( 5x^2 - 6x - 11 = x^2 - 9 \) rounded to the nearest tenth are:
\[ x \approx 1.8 \text{ and } x \approx -0.3. \]