**Solve:**

**\(3x^{2}\ +\ 16x = −5\)**

Write the equation in standard form.

**\(3x^{2}\ +\ 16\ +\ 5 = 0\)**

Solve using the quadratic formula.

**\(a = 3, b = 16, c = 5\)**

\(x = \frac{−b\ \pm\ \sqrt{b^{2}\ −\ 4ac}}{2a}\)

\(= \frac{−16\ \pm\ \sqrt{16^{2}\ −\ 4(3)(5)}}{2(3)}\)

\(= \frac{−16\ \pm\ \sqrt{256\ −\ 60}}{6}\)

\(= \frac{16\ \pm\ \sqrt{196}}{6}\)

\(= \frac{16\ \pm\ 14}{6}\)

\(x= \frac{−16\ +\ 14}{6}=−\frac{1}{3}\ or\ x = \frac{−16\ −\ 14}{6} = −5\)

**The solutions are –5 and ****\(−\frac{1}{3}\)**