Solve giving the answer correct to three significant figures: $\;$ $x - \dfrac{18}{x} = 6$
Given quadratic equation: $\;$ $x - \dfrac{18}{x} = 6$
i.e. $\;$ $x^2 - 6x - 18 = 0$
Comparing with the standard quadratic equation $\;$ $ax^2 + bx + c = 0$ $\;$ gives
$a = 1, \; b = -6, \; c = -18$
Solution of the quadratic equation is (by quadratic formula)
$\begin{aligned}
x & = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\\\
& = \dfrac{6 \pm \sqrt{\left(-6\right)^2 - 4 \times 1 \times \left(-18\right)}}{2 \times 1} \\\\
& = \dfrac{6 \pm \sqrt{36 + 72}}{2} \\\\
& = \dfrac{6 \pm \sqrt{108}}{2} \\\\
& = \dfrac{6 \pm 10.392}{2} \\\\
& = \dfrac{16.392}{2} \; or \; \dfrac{-4.392}{2} \\\\
& = 8.196 \; or \; -2.196
\end{aligned}$
i.e. $\;$ $x = 8.20$ $\;$ or $\;$ $x = -2.20$ $\;\;$ [correct to 3 significant figures]