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