# Chapter 10 - Quadratic Equations - Chapter 10 Review Problem Set - Page 467: 7

{$\frac{-7-\sqrt {61}}{6},\frac{-7+\sqrt {61}}{6}$}

#### Work Step by Step

Step 1: Comparing $3a^{2}+7a-1=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find: *Note, the "a" in the standard form of a quadratic formula is a constant while the "a" in the quadratic equation is a variable. Step 2: The quadratic formula to be used is: $a=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$ Step 3: Substituting the values of a, b and c in the formula: $a=\frac{-(7) \pm \sqrt {(7)^{2}-4(3)(-1)}}{2(3)}$ Step 4: $a=\frac{-7 \pm \sqrt {49+12}}{6}$ Step 5: $a=\frac{-7 \pm \sqrt {61}}{6}$ Step 6: $a=\frac{-7-\sqrt {61}}{6}$ or $a=\frac{-7+\sqrt {61}}{6}$ Step 7: Therefore, the solution set is {$\frac{-7-\sqrt {61}}{6},\frac{-7+\sqrt {61}}{6}$}.

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