Answer
{$\frac{-1-\sqrt {21}}{10},\frac{-1+\sqrt {21}}{10}$}
Work Step by Step
Step 1: Comparing $5x^{2}+x-1=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find:
$a=5$, $b=1$ and $c=-1$
Step 2: The quadratic formula is:,
$x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$
Step 3: Substituting the values of a, b, and c in the formula:
$x=\frac{-(1) \pm \sqrt {(1)^{2}-4(5)(-1)}}{2(5)}$
Step 4: $x=\frac{-1 \pm \sqrt {1+20}}{10}$
Step 5: $x=\frac{-1 \pm \sqrt {21}}{10}$
Step 6: $x=\frac{-1-\sqrt {21}}{10}$ or $x=\frac{-1+\sqrt {21}}{10}$
Step 7: Therefore, the solution set is {$\frac{-1-\sqrt {21}}{10},\frac{-1+\sqrt {21}}{10}$}.
