Solve the equation: \( 3(x+2)^{2}+29(x+2)+40=0 \)

Let \( y = x + 2 \), then the equation transforms into: \[ 3y^2 + 29y + 40 = 0. \] Now, we can apply the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \] where \( a = 3 \), \( b = 29 \), and \( c = 40 \). Calculating the discriminant: \[ b^2 - 4ac = 29^2 - 4 \times 3 \times 40 = 841 - 480 = 361. \] Now substitute back into the quadratic formula: \[ y = \frac{-29 \pm \sqrt{361}}{2 \times 3} = \frac{-29 \pm 19}{6}. \] This results in two potential solutions: 1. \( y = \frac{-29 + 19}{6} = \frac{-10}{6} = -\frac{5}{3} \). 2. \( y = \frac{-29 - 19}{6} = \frac{-48}{6} = -8 \). Now reverting back to \( x \): 1. For \( y = -\frac{5}{3} \): \[ x + 2 = -\frac{5}{3} \Rightarrow x = -\frac{5}{3} - 2 = -\frac{5}{3} - \frac{6}{3} = -\frac{11}{3}. \] 2. For \( y = -8 \): \[ x + 2 = -8 \Rightarrow x = -8 - 2 = -10. \] Thus, the solutions to the original equation are: \[ x = -\frac{11}{3} \quad \text{and} \quad x = -10. \]