Solve the equation by the method of your choice. \[ (5 x+2)(x+1)=1 \] The solution set is \{ \} (Simplify your answer. Type an exact answer, using radicals and \( i \) as needed. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.)

To solve the equation \((5x + 2)(x + 1) = 1\), start by expanding the left side: \[ 5x^2 + 5x + 2x + 2 = 1 \] Simplifying this gives: \[ 5x^2 + 7x + 2 = 1 \] Now, move the \(1\) to the left side: \[ 5x^2 + 7x + 1 = 0 \] Next, we can use the quadratic formula, which states that for any equation of the form \(ax^2 + bx + c = 0\), the solutions are given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In our case, \(a = 5\), \(b = 7\), and \(c = 1\). Plugging in these values, we first calculate the discriminant: \[ b^2 - 4ac = 7^2 - 4 \times 5 \times 1 = 49 - 20 = 29 \] Now we can find the solutions: \[ x = \frac{-7 \pm \sqrt{29}}{2 \times 5} \] This simplifies to: \[ x = \frac{-7 \pm \sqrt{29}}{10} \] Thus, the solution set is: \[ \left\{ \frac{-7 + \sqrt{29}}{10}, \frac{-7 - \sqrt{29}}{10} \right\} \]