Answer
$x = - \frac{118}{16} \frac{}{+} \frac{\sqrt {119}i}{8}$
Work Step by Step
$$x^{\frac{1}{2}} - 2x^{\frac{1}{4}} - 15 = 0$$
$x^{\frac{1}{2}} - 2(x^{\frac{1}{2}})^{2} - 15 = 0$; where, if we consider $y=x^{\frac{1}{2}}$, this becomes:
$y - 2y^{2} - 15= 0$; where we can use the quadratic formula $x = \frac{-b \frac{+}{} \sqrt {b^{2} - 4ac}}{2a}$:
$y = \frac{-(1) \frac{+}{} \sqrt {(1)^{2} - 4(-2)(-15)}}{2(-2)} = \frac{-1 \frac{+}{} \sqrt {1 - 120}}{-4} = \frac{-1 \frac{+}{} \sqrt {-119}}{-4}$
Now, solving once again for $x$:
$y = x^{\frac{1}{2}} = \frac{1}{4} \frac{+}{} (-\frac{1}{4}\sqrt {-119})$
$[x^{\frac{1}{2}}]^{2} = [\frac{1}{4} \frac{+}{} (-\frac{1}{4}\sqrt {-119})]^{2}$
$x = [(\frac{1}{4})^{2} + (-\frac{1}{4}\sqrt {-119})^{2} \frac{+}{} (2)(\frac{1}{4})(-\frac{1}{4}\sqrt {-119})]$
$x = \frac{1}{16} - \frac{119}{16} \frac{}{+} \frac{\sqrt {119}i}{8} = - \frac{118}{16} \frac{}{+} \frac{\sqrt {119}i}{8}$
