## Elementary Algebra

Step 1: We write $-2x^{2}-3x-1=0$ as $2x^{2}+3x+1=0$. Comparing $2x^{2}+3x+1=0$ to the standard form of a quadratic equation $ax^{2}+bx+c=0$, we obtain: $a=2$, $b=3$ 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: $x=\frac{-(3) \pm \sqrt {(3)^{2}-4(2)(1)}}{2(2)}$ Step 4: $x=\frac{-3 \pm \sqrt {9-8}}{4}$ Step 5: $x=\frac{-3 \pm \sqrt {1}}{4}$ Step 6: $x=\frac{-3 \pm 1}{4}$ Step 7: $x=\frac{-3-1}{4}$ or $x=\frac{-3+1}{4}$ Step 8: $x=\frac{-4}{4}$ or $x=\frac{-2}{4}$ Step 9: $x=-1$ or $x=\frac{-1}{2}$ Step 10: Therefore, the solution set is {$-\frac{1}{2},-1$}. Therefore, it is true that the solution set consists of two rational numbers.