Answer
{$-\frac{3}{4},1$}
Work Step by Step
Step 1: Comparing $4x^{2}-x-3=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we obtain:
$a=4$, $b=-1$, and $c=-3$
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, we obtain:
$x=\frac{-(-1) \pm \sqrt {(-1)^{2}-4(4)(-3)}}{2(4)}$
Step 4: $x=\frac{1 \pm \sqrt {1+48}}{8}$
Step 5: $x=\frac{1 \pm \sqrt {49}}{8}$
Step 6: $x=\frac{1 \pm 7}{8}$
Step 7: $x=\frac{1 + 7}{8}$ or $x=\frac{1 - 7}{8}$
Step 8: $x=\frac{8}{8}$ or $x=\frac{-6}{8}$
Step 9: $x=1$ or $x=\frac{-3}{4}$
Step 10: Therefore, the solution set is {$-\frac{3}{4},1$}.
