## Elementary Algebra

Published by Cengage Learning

# Chapter 10 - Quadratic Equations - 10.3 - Quadratic Formula - Problem Set 10.3: 36

#### Answer

{$\frac{-3-\sqrt {13}}{4},\frac{-3+\sqrt {13}}{4}$}

#### Work Step by Step

Step 1: Comparing $4n^{2}+6n-1=0$ to the standard form of a quadratic equation, $an^{2}+bn+c=0$, we find: $a=4$, $b=6$ and $c=-1$ Step 2: The quadratic formula is: $n=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$ Step 3: Substituting the values of a,b and c in the formula: $n=\frac{-(6) \pm \sqrt {(6)^{2}-4(4)(-1)}}{2(4)}$ Step 4: $n=\frac{-6 \pm \sqrt {36+16}}{8}$ Step 5: $n=\frac{-6 \pm \sqrt {52}}{8}$ Step 6: $n=\frac{-6 \pm \sqrt {4\times13}}{8}$ Step 7: $n=\frac{-6 \pm 2\sqrt {13}}{8}$ Step 8: $n=\frac{2(-3 \pm \sqrt {13})}{8}$ Step 9: $n=\frac{(-3 \pm \sqrt {13})}{4}$ Step 10: $n=\frac{-3-\sqrt {13}}{4}$ or $n=\frac{-3+\sqrt {13}}{4}$ Step 11: Therefore, the solution set is {$\frac{-3-\sqrt {13}}{4},\frac{-3+\sqrt {13}}{4}$}.

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