Answer
{$\frac{-5-\sqrt {137}}{14},\frac{-5+\sqrt {137}}{14}$}
Work Step by Step
Step 1: Comparing $7x^{2}+5x-4=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find:
$a=7$, $b=5$ 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:
$x=\frac{-(5) \pm \sqrt {(5)^{2}-4(7)(-4)}}{2(7)}$
Step 4: $x=\frac{-5 \pm \sqrt {25+112}}{14}$
Step 5: $x=\frac{-5 \pm \sqrt {137}}{14}$
Step 6: $x=\frac{-5-\sqrt {137}}{14}$ or $x=\frac{-5+\sqrt {137}}{14}$
Step 7: Therefore, the solution set is {$\frac{-5-\sqrt {137}}{14},\frac{-5+\sqrt {137}}{14}$}.
