Solve the following quadratic equation:

Given,

⇒ 3x² + 6x - 4 = 0

Comparing equation 3x² + 6x - 4 = 0 with ax² + bx + c = 0, we get :

a = 3, b = 6 and c = -4.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow x = \frac{-(6) \pm \sqrt{(6)^2 - 4 \times 3 \times (-4)}}{2 \times 3} \\[1em] = \frac{-6 \pm \sqrt{36 + 48}}{6} \\[1em] = \frac{-6 \pm \sqrt{84}}{6} \\[1em] = \frac{-6 \pm \sqrt{21 \times 4}}{6} \\[1em] = \frac{-6 \pm 2\sqrt{21}}{6} \\[1em] = \frac{-6 + 2\sqrt{21}}{6} \text{ or } \frac{-6 - 2\sqrt{21}}{6} \\[1em] = \frac{2(-3 + \sqrt{21})}{6} \text{ or } \frac{2(-3 - \sqrt{21})}{6} \\[1em] = \frac{-3 + \sqrt{21}}{3} \text{ or } \frac{-3 - \sqrt{21}}{3} \\[1em] = \frac{-3 + 4.58}{3} \text{ or } \frac{-3 - 4.58}{3} \\[1em] = \frac{1.58}{3} \text{ or } -\frac{7.58}{3} \\[1em] = 0.53 \text{ or } -2.53 \\[1em]\end{aligned}$$

Hence, x = 0.53 or -2.53.