Answer
{$-4,1$}
Work Step by Step
Step 1: Comparing $x^{2}+3x-4=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we obtain:
$a=1$, $b=3$ and $c=-4$
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:
$x=\frac{-(3) \pm \sqrt {(3)^{2}-4(1)(-4)}}{2(1)}$
Step 4: $x=\frac{-3 \pm \sqrt {9+16}}{2}$
Step 5: $x=\frac{-3 \pm \sqrt {25}}{2}$
Step 6: $x=\frac{-3 \pm 5}{2}$
Step 7: $x=\frac{-3+5}{2}$ or $x=\frac{-3-5}{2}$
Step 8: $x=\frac{2}{2}$ or $x=\frac{-8}{2}$
Step 9: $x=1$ or $x=-4$
Step 10: Therefore, the solution set is {$-4,1$}.
