1. We are asked to find the non-real solutions to the quadratic equation $x^2 + 4x + 5 = 0$ using the quadratic formula. 2. Recall the quadratic formula for solutions of $ax^2 + bx + c = 0$ is: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 3. For our equation, $a = 1$, $b = 4$, and $c = 5$. 4. Calculate the discriminant: $$\Delta = b^2 - 4ac = 4^2 - 4 \times 1 \times 5 = 16 - 20 = -4$$ Since the discriminant is negative, solutions are complex (non-real). 5. Substitute into the quadratic formula: $$x = \frac{-4 \pm \sqrt{-4}}{2 \times 1} = \frac{-4 \pm 2i}{2}$$ 6. Simplify: $$x = -2 \pm i$$ 7. Thus, the non-real solutions are $x = -2 + i$ and $x = -2 - i$. **Final answer:** $x = -2 \pm i$