Solve the following equation using quadratic formula:

x² + 7x = 7

Given,

⇒ x² + 7x - 7 = 0

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

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

By formula,

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

Substituting values we get :

$\Rightarrow x = \dfrac{-(7) \pm \sqrt{(7)^2 - 4 \times (1) \times (-7)}}{2 \times (1)} \\[1em] = \dfrac{-7 \pm \sqrt{49 + 28}}{2} \\[1em] = \dfrac{-7 \pm \sqrt{77}}{2} \\[1em] = \dfrac{-7 + \sqrt{77}}{2} \text{ or } \dfrac{-7 - \sqrt{77}}{2} \\[1em] = \dfrac{-7 + 8.77}{2} \text{ or } \dfrac{-7 - 8.77}{2} \\[1em] = \dfrac{1.77}{2} \text{ or } \dfrac{-15.77}{2} \\[1em] = 0.89 \text{ or } -7.89.$

Hence, x = {0.89, -7.89}.