Solve the following equations by using formula:

$x \Big(3x + \dfrac{1}{2} \Big)$ = 6

The given equation is $x \Big(3x + \dfrac{1}{2} \Big)$ = 6

$\Rightarrow 3x^2 + \dfrac{1}{2}x = 6 \\[1em] \Rightarrow 6x^2 + x = 12 \\[1em] \Rightarrow 6x^2 + x - 12 = 0$

Comparing it with ax² + bx + c = 0
a= 6, b = 1, c = -12

By using the formula , x = $\dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$\Rightarrow \dfrac{-(1) ± \sqrt{(1)^2 - 4 \times 6 \times -12 }}{2 \times 6} \\[1em] \Rightarrow \dfrac{-1 ± \sqrt{1 + 288}}{12} \\[1em] \Rightarrow \dfrac{-1 + \sqrt{289}}{12} \text{ or } \dfrac{-1 - \sqrt{289}}{12} \\[1em] \Rightarrow \dfrac{-1 + 17}{12} \text{ or } \dfrac{-1 - 17}{12} \\[1em] \Rightarrow \dfrac{16}{12} \text{ or } -\dfrac{18}{12} \\[1em] \Rightarrow \dfrac{4}{3} \text{ or } -\dfrac{3}{2}$

Hence, roots of the equation are $\dfrac{4}{3} , -\dfrac{3}{2}$.