Solving a Quadratic Equation

To solve this quadratic equation, we’ll use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Where $a$, $b$, and $c$ are the coefficients in the standard form of a quadratic equation: $ax^2 + bx + c = 0$.

For our equation $x^2 - 5x + 6 = 0$:

• $a = 1$
• $b = -5$
• $c = 6$

Let’s substitute these values into the quadratic formula:

$x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(6)}}{2(1)}$

$x = \frac{5 \pm \sqrt{25 - 24}}{2}$

$x = \frac{5 \pm \sqrt{1}}{2}$

$x = \frac{5 \pm 1}{2}$

This gives us two solutions:

$x_1 = \frac{5 + 1}{2} = \frac{6}{2} = 3$

$x_2 = \frac{5 - 1}{2} = \frac{4}{2} = 2$

Therefore, the solutions to the equation $x^2 - 5x + 6 = 0$ are:

$x = 3 \text{ or } x = 2$

You can verify these solutions by substituting them back into the original equation:

For $x = 3$: $(3)^2 - 5(3) + 6 = 9 - 15 + 6 = 0$

For $x = 2$: $(2)^2 - 5(2) + 6 = 4 - 10 + 6 = 0$

Both solutions satisfy the equation, confirming our answer.