Solve the equation \( 4 x^{2}+35 x+49=0 \) by factoring. The solutions are \( x_{1}=\square \) and \( x_{2}=\square \) with \( x_{1} \leq x_{2} \). Note: You can earn partial credit on this problem.

To solve the equation \( 4x^2 + 35x + 49 = 0 \) by factoring, we start by looking for two numbers that multiply to \( 4 \cdot 49 = 196 \) and add up to \( 35 \). The two numbers that meet these criteria are \( 28 \) and \( 7 \). We can rewrite the equation using these numbers: \[ 4x^2 + 28x + 7x + 49 = 0 \] Next, we group the terms: \[ (4x^2 + 28x) + (7x + 49) = 0 \] Factoring out the common factors from each group, we get: \[ 4x(x + 7) + 7(x + 7) = 0 \] Now we can factor out the common binomial \( (x + 7) \): \[ (4x + 7)(x + 7) = 0 \] Now we set each factor equal to zero: 1. \( 4x + 7 = 0 \) leads to \( 4x = -7 \) or \( x = -\frac{7}{4} \) 2. \( x + 7 = 0 \) leads to \( x = -7 \) Thus, the solutions are \( x_1 = -7 \) and \( x_2 = -\frac{7}{4} \) with \( x_1 \leq x_2 \). So, we have: \( x_{1} = -7 \) and \( x_{2} = -\frac{7}{4} \).