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, which is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our equation, the coefficients are:
- \(a = 2\)
- \(b = -5\)
- \(c = -6\)
Now, we will calculate the discriminant \(b^2 - 4ac\):
\[ b^2 - 4ac = (-5)^2 - 4 \cdot 2 \cdot (-6) = 25 + 48 = 73 \]
Since the discriminant is positive, we will have two real solutions. We can now substitute the values into the quadratic formula:
\[ x = \frac{-(-5) \pm \sqrt{73}}{2 \cdot 2} = \frac{5 \pm \sqrt{73}}{4} \]
Next, we calculate the two possible solutions:
- For \(x_1\): \[ x_1 = \frac{5 + \sqrt{73}}{4} \]
Calculating \(\sqrt{73} \approx 8.544\): \[ x_1 \approx \frac{5 + 8.544}{4} = \frac{13.544}{4} \approx 3.386 \]
- For \(x_2\): \[ x_2 = \frac{5 - \sqrt{73}}{4} \] Calculating again \(\sqrt{73} \approx 8.544\): \[ x_2 \approx \frac{5 - 8.544}{4} = \frac{-3.544}{4} \approx -0.886 \]
Now, we round the solutions to the nearest hundredth:
- \(x_1 \approx 3.39\)
- \(x_2 \approx -0.89\)
Thus, the rounded solutions to the equation \(2x^2 - 5x = 6\) are approximately \(3.39\) and \(-0.89\).