# Chapter 10 - Quadratic Equations - 10.3 - Quadratic Formula - Problem Set 10.3 - Page 453: 41

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.

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