To determine the values of \( a \) for which the quadratic equation \( x^2 + 4x + a^2 - 3a = 0 \) has real roots, we need to ensure that the discriminant is non-negative. The general form of a quadratic equation is \( ax^2 + bx + c = 0 \), and the discriminant \( D \) is given by: \[ D = b^2 - 4ac \] For the equation \( x^2 + 4x + a^2 - 3a = 0 \): - \( a = 1 \) - \( b = 4 \) - \( c = a^2 - 3a \) The discriminant is: \[ D = 4^2 - 4 \cdot 1 \cdot (a^2 - 3a) = 16 - 4(a^2 - 3a) \] Simplify the discriminant: \[ D = 16 - 4a^2 + 12a = -4a^2 + 12a + 16 \] For the roots to be real, \( D \geq 0 \): \[ -4a^2 + 12a + 16 \geq 0 \] Divide the entire inequality by -4 (remembering to reverse the inequality sign): \[ a^2 - 3a - 4 \leq 0 \] Solve the quadratic inequality \( a^2 - 3a - 4 \leq 0 \): First, find the roots of the equation \( a^2 - 3a - 4 = 0 \): \[ a = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm 5}{2} \] \[ a = 4 \quad \text{or} \quad a = -1 \] The quadratic \( a^2 - 3a - 4 \) is a parabola opening upwards. Therefore, the inequality \( a^2 - 3a - 4 \leq 0 \) holds for \( a \) between the roots: \[ -1 \leq a \leq 4 \] Final Answer: \[ \boxed{D} \]