GGoogle Classroom \( \square \quad \) Molve for \( x \). Enter the solutions from least to greatest. \( 2 x^{2}-24 x+54=0 \) lesser \( x=\square \) greater \( x=\square \)

To solve the quadratic equation \(2x^2 - 24x + 54 = 0\), we can first simplify it by dividing the entire equation by 2, giving us \(x^2 - 12x + 27 = 0\). Next, we apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -12\), and \(c = 27\). Calculating the discriminant: \[ b^2 - 4ac = (-12)^2 - 4(1)(27) = 144 - 108 = 36 \] Now we use the quadratic formula: \[ x = \frac{12 \pm \sqrt{36}}{2 \cdot 1} = \frac{12 \pm 6}{2} \] This gives us two solutions: \[ x_1 = \frac{18}{2} = 9 \quad \text{and} \quad x_2 = \frac{6}{2} = 3 \] Thus, the solutions from least to greatest are: lesser \( x=3 \) greater \( x=9 \)