## Algebra and Trigonometry 10th Edition

$\frac{1\pm i\sqrt {11} }{8}$
The quadratic formula says $x = \frac{ - b \pm \sqrt {b^2 - 4ac} }{2a}$ when $ax^2 + bx + c = 0$. We know that $i^2=-1$. Hence here $t = \frac{ - (-4) \pm \sqrt {(-4)^2 - 4\cdot16\cdot3} }{2\cdot16}= \frac{4\pm \sqrt {16 - 192} }{32}= \frac{4\pm \sqrt {-176} }{32}= \frac{4\pm 4i\sqrt {11} }{32}= \frac{1\pm i\sqrt {11} }{8}$