Solve the equation: \[ (4 a+3)^{2}+7(4 a+3)+10=0 \]

To solve the equation \( (4a+3)^2 + 7(4a+3) + 10 = 0 \), we can make a substitution to simplify it. Let \( x = 4a + 3 \). Then the equation transforms into: \[ x^2 + 7x + 10 = 0 \] Next, we can factor this quadratic equation. We need two numbers that multiply to \( 10 \) and add up to \( 7 \). Those numbers are \( 5 \) and \( 2 \). Thus, we can factor the equation as follows: \[ (x + 5)(x + 2) = 0 \] Setting each factor to zero gives us: \[ x + 5 = 0 \quad \Rightarrow \quad x = -5 \] \[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \] Now we substitute back \( x = 4a + 3 \) into these results. 1. For \( x = -5 \): \[ 4a + 3 = -5 \\ 4a = -5 - 3 \\ 4a = -8 \\ a = -2 \] 2. For \( x = -2 \): \[ 4a + 3 = -2 \\ 4a = -2 - 3 \\ 4a = -5 \\ a = -\frac{5}{4} \] Thus, the solutions to the original equation are: \[ \boxed{-2} \quad \text{and} \quad \boxed{-\frac{5}{4}} \]