1.3 Given: \( f(x)=x^{2}-5 x+c \) Determine the value(s) of \( c \) if it is given that the solutions of \( f(x)=0 \) are \( \frac{5 \pm \sqrt{41}}{2} \).

To solve for \( c \) given the function \( f(x) = x^2 - 5x + c \) and the solutions of \( f(x) = 0 \) are \( \frac{5 \pm \sqrt{41}}{2} \), we can use the quadratic formula, which states that for a quadratic equation \( ax^2 + bx + c = 0 \), the solutions are given by \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \] In our case, \( a = 1 \), \( b = -5 \), and \( c \) is the unknown. Thus, replacing these values into the formula gives: \[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(c)}}{2(1)} = \frac{5 \pm \sqrt{25 - 4c}}{2}. \] We are given that the solutions are \( \frac{5 \pm \sqrt{41}}{2} \). Equating the expressions for the square roots from both sides, we find: \[ \sqrt{25 - 4c} = \sqrt{41}. \] Next, we square both sides to eliminate the square root: \[ 25 - 4c = 41. \] Now, solving for \( c \): \[ -4c = 41 - 25, \] \[ -4c = 16, \] \[ c = -4. \] Thus, the value of \( c \) is \( \boxed{-4} \).