To solve the equation \( 2x^2 + 3x - 6 = 5 \), we first move all terms to one side of the equation:
\[ 2x^2 + 3x - 6 - 5 = 0 \]
This simplifies to:
\[ 2x^2 + 3x - 11 = 0 \]
Now, we can use the quadratic formula, which is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our equation, \( a = 2 \), \( b = 3 \), and \( c = -11 \).
First, we can calculate the discriminant \( b^2 - 4ac \):
\[ b^2 - 4ac = 3^2 - 4(2)(-11) = 9 + 88 = 97 \]
Now substituting into the quadratic formula:
\[ x = \frac{-3 \pm \sqrt{97}}{2(2)} = \frac{-3 \pm \sqrt{97}}{4} \]
Thus, the solutions are:
\[ x = \frac{-3 + \sqrt{97}}{4} \quad \text{and} \quad x = \frac{-3 - \sqrt{97}}{4} \]
So, the final solutions are:
\[ x \approx 1.7 \quad \text{and} \quad x \approx -2.7 \]
These can be left in the exact form:
\[ x = \frac{-3 + \sqrt{97}}{4}, \quad x = \frac{-3 - \sqrt{97}}{4} \]
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