Step 1: Factor the quadratic Find two numbers that multiply to 6 and add to -5: $-2$ and $-3$

$x^2 - 5x + 6 = (x - 2)(x - 3) = 0$

Step 2: Set each factor equal to zero $x - 2 = 0 \quad \text{or} \quad x - 3 = 0$

Step 3: Solve each equation $x = 2 \quad \text{or} \quad x = 3$

Answer: $x = 2$ or $x = 3$

Identify: $a = 1$, $b = 6$, $c = 2$

Use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Step 1: Calculate the discriminant $b^2 - 4ac = (6)^2 - 4(1)(2) = 36 - 8 = 28$

Step 2: Substitute into the formula $x = \frac{-6 \pm \sqrt{28}}{2}$

Step 3: Simplify $\sqrt{28}$ $\sqrt{28} = \sqrt{4 \cdot 7} = 2\sqrt{7}$

$x = \frac{-6 \pm 2\sqrt{7}}{2} = \frac{2(-3 \pm \sqrt{7})}{2} = -3 \pm \sqrt{7}$

Answer: $x = -3 + \sqrt{7}$ or $x = -3 - \sqrt{7}$

Use the discriminant to determine the number of solutions: $\text{Discriminant} = b^2 - 4ac$

For $x^2 + 4x + 5 = 0$: $a = 1$, $b = 4$, $c = 5$

$b^2 - 4ac = (4)^2 - 4(1)(5) = 16 - 20 = -4$

Since the discriminant is negative, the equation has no real solutions (it has 2 complex solutions).

Answer: No real solutions