Answer:
4.
[tex]$x=\frac{2}{3} \pm \frac{1}{6}i \sqrt{158}$[/tex]
Step-by-step explanation:
[tex]18x^2-24x+87=0[/tex]
Using Quadratic Formula
[tex]$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$[/tex]
[tex]$x=\frac{-\left(-24\right)\pm \sqrt{\left(-24\right)^2-4\cdot \:18\cdot \:87}}{2\cdot \:18}$[/tex]
Solving the discriminant:
[tex]\Delta = \left(-24\right)^2-4\cdot \:18\cdot \:87\\\Delta = 576-6264\\ \Delta=-5688[/tex]
[tex]$x= \frac{24 \pm \sqrt{-5688} }{36} $[/tex]
[tex]$x= \frac{24 \pm \sqrt{5688i} }{36} $[/tex]
Once [tex]\sqrt{5688} =\sqrt{2^3\cdot \:3^2\cdot \:79}=6\sqrt{158}[/tex]
[tex]$x=\frac{24\pm6\sqrt{158}i}{36}$[/tex]
Dividing the denominator and numerator by 6
[tex]$x=\frac{4\pm \sqrt{158}i}{6}$[/tex]
Now rewrite it:
[tex]$x=\frac{4}{6} \pm i\frac{\sqrt{158}}{6}$[/tex]
[tex]$x=\frac{2}{3} \pm i \frac{\sqrt{158}}{6}$[/tex]
or
[tex]$x=\frac{2}{3} \pm \frac{1}{6}i \sqrt{158}$[/tex]
