Answer
The equation does not have any real solutions
Work Step by Step
Step 1: Comparing $3t^{2}+6t+5=0$ to the standard form of a quadratic equation, $at^{2}+bt+c=0$, we find:
$a=3$, $b=6$ and $c=5$
Step 2: The quadratic formula is:
$y=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$
Step 3: Substituting the values of a, b, and c in the formula:
$y=\frac{-(6) \pm \sqrt {(6)^{2}-4(3)(5)}}{2(3)}$
Step 4: $y=\frac{-6 \pm \sqrt {36-60}}{6}$
Step 5: $y=\frac{-6 \pm \sqrt {-24}}{6}$
Step 6: Since $\sqrt {-24}$ is not a real number, the equation does not have any real solutions.
