Solve the following equation for x and give your answer correct to two decimal places :

4x + $\dfrac{6}{x}$ + 13 = 0

Given,

$\Rightarrow 4x + \dfrac{6}{x} + 13 = 0 \\[1em] \Rightarrow \dfrac{4x^2 + 6 + 13x}{x} = 0 \\[1em] \Rightarrow 4x^2 + 13x + 6 = 0$

Comparing 4x² + 13x + 6 = 0 with ax² + bx + c = 0 we get,

a = 4, b = 13 and c = 6.

We know that,

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

Substituting values of a, b and c in above equation we get,

$\Rightarrow x = \dfrac{-(13) \pm \sqrt{(-13)^2 - 4.(4).(6)}}{2(4)} \\[1em] = \dfrac{-13 \pm \sqrt{169 - 96}}{8} \\[1em] = \dfrac{-13 \pm \sqrt{73}}{8} \\[1em] = \dfrac{-13 \pm 8.54}{8} \\[1em] = \dfrac{-13 + 8.54}{8} \text{ and } \dfrac{-13 - 8.54}{8} \\[1em] = \dfrac{-4.46}{8} \text{ and } \dfrac{-21.54}{8} \\[1em] = -0.56 \text{ and } -2.69$

Hence, x = -0.56 and -2.69