Solve the equation by any method. \[ 7 x^{2}+24 x+20=0 \]

To solve the quadratic equation \( 7x^2 + 24x + 20 = 0 \), we can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 7 \), \( b = 24 \), and \( c = 20 \). Let's calculate the discriminant \( b^2 - 4ac \): \[ b^2 = 24^2 = 576 \] \[ 4ac = 4 \cdot 7 \cdot 20 = 560 \] \[ b^2 - 4ac = 576 - 560 = 16 \] Since the discriminant is positive, there are two distinct real roots. Now substituting into the quadratic formula: \[ x = \frac{-24 \pm \sqrt{16}}{2 \cdot 7} \] \[ x = \frac{-24 \pm 4}{14} \] Calculating the two possible values for \( x \): 1. \( x = \frac{-24 + 4}{14} = \frac{-20}{14} = -\frac{10}{7} \) 2. \( x = \frac{-24 - 4}{14} = \frac{-28}{14} = -2 \) Thus, the solutions to the equation are: \[ x = -\frac{10}{7} \quad \text{and} \quad x = -2 \]