Answer
{$-0.90,3.10$}
Work Step by Step
Step 1: Comparing $5x^{2}-11x-14=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we obtain:
$a=5$, $b=-11$, and $c=-14$
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:
$x=\frac{-(-11) \pm \sqrt {(-11)^{2}-4(5)(-14)}}{2(5)}$
Step 4: $x=\frac{11 \pm \sqrt {121+280}}{10}$
Step 5: $x=\frac{11 \pm \sqrt {401}}{10}$
Step 6: $x=\frac{11 \pm 20.025}{10}$
Step 7: $x=\frac{11+20.025}{10}$ or $x=\frac{11-20.025}{10}$
Step 8: $x=\frac{31.025}{10}$ or $x=\frac{-9.025}{10}$
Step 9: $x=3.1025\approx3.10$ or $x=-0.9025\approx-0.90$
Step 10: Therefore, the solution set is {$-0.90,3.10$}.
