Answer
The equation does not have any real solutions
Work Step by Step
Step 1: Comparing $4t^{2}+5t+3=0$ to the standard form of a quadratic equation, $at^{2}+bt+c=0$, we find:
$a=4$, $b=5$ and $c=3$
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:
$y=\frac{-(5) \pm \sqrt {(5)^{2}-4(4)(3)}}{2(4)}$
Step 4: $y=\frac{-5 \pm \sqrt {25-48}}{8}$
Step 5: $y=\frac{-5 \pm \sqrt {-23}}{8}$
Step 6: Since $\sqrt {-23}$ is not a real number, the equation does not have any real solutions.
