To solve the quadratic equation \(2x^2 - 5x = 6\), we first rearrange it into standard form:
\[2x^2 - 5x - 6 = 0.\]
Next, we can use the quadratic formula to find the solutions:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]
where \(a = 2\), \(b = -5\), and \(c = -6\).
First, we need to calculate the discriminant (\(b^2 - 4ac\)):
\[ b^2 - 4ac = (-5)^2 - 4(2)(-6) = 25 + 48 = 73. \]
Now we can use the quadratic formula:
\[ x = \frac{-(-5) \pm \sqrt{73}}{2(2)} = \frac{5 \pm \sqrt{73}}{4}. \]
Calculating the two possible values for \(x\):
\[ x_1 = \frac{5 + \sqrt{73}}{4} \quad \text{and} \quad x_2 = \frac{5 - \sqrt{73}}{4}. \]
Now we need to calculate these values approximately.
- Calculating \(x_1\):
\[ \sqrt{73} \approx 8.54 \] \[ x_1 \approx \frac{5 + 8.54}{4} \approx \frac{13.54}{4} \approx 3.385. \]
- Calculating \(x_2\):
\[ x_2 \approx \frac{5 - 8.54}{4} \approx \frac{-3.54}{4} \approx -0.885. \]
Rounding these to the nearest hundredth, we have:
\[ x_1 \approx 3.39 \quad \text{and} \quad x_2 \approx -0.89. \]
Thus, the solutions to the equation \(2x^2 - 5x = 6\) rounded to the nearest hundredth are approximately \(3.39\) and \(-0.89\).