Solve the following equation using quadratic formula:

x² - 7x + 3 = 0

Given,

⇒ x² - 7x + 3 = 0

Comparing equation x² - 7x + 3 = 0 with ax² + bx + c = 0, we get :

a = 1, b = -7 and c = 3.

By formula,

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

Substituting values we get :

$\Rightarrow x = \dfrac{-(-7) \pm \sqrt{(-7)^2 - 4 \times (1) \times (3)}}{2 \times (1)} \\[1em] = \dfrac{7 \pm \sqrt{49 - 12}}{2} \\[1em] = \dfrac{7 \pm \sqrt{37}}{2} \\[1em] = \dfrac{7 + \sqrt{37}}{2} \text{ or } \dfrac{7 - \sqrt{37}}{2} \\[1em] = \dfrac{7 + 6.08}{2} \text{ or } \dfrac{7 - 6.08}{2} \\[1em] = \dfrac{13.08}{2} \text{ or } \dfrac{0.92}{2} \\[1em] = 6.54 \text{ or } 0.46$

Hence, x = {6.54, 0.46}.