## Algebra and Trigonometry 10th Edition

$\frac{6\pm i\sqrt {34} }{14}$
The quadratic formula says $x = \frac{ - b \pm \sqrt {b^2 - 4ac} }{2a}$ when $ax^2 + bx + c = 0$. We know that $i^2=-1$. If I multiply the equation by $16$, the coefficients will be $14,-12,5$. Hence here $x = \frac{ - (-12) \pm \sqrt {(-12)^2 - 4\cdot14\cdot5} }{2\cdot14}= \frac{12\pm \sqrt {144 - 280} }{28}= \frac{12\pm \sqrt {-136} }{28}= \frac{12\pm 2i\sqrt {34} }{28}= \frac{6\pm i\sqrt {34} }{14}$