Answer
{$\frac{-5 - \sqrt {73}}{4},\frac{-5+\sqrt {73}}{4}$}
Work Step by Step
Step 1: Comparing $2x^{2}+5x-6=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find:
$a=2$, $b=5$ and $c=-6$
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:
$x=\frac{-(5) \pm \sqrt {(5)^{2}-4(2)(-6)}}{2(2)}$
Step 4: $x=\frac{-5 \pm \sqrt {25+48}}{4}$
Step 5: $x=\frac{-5 \pm \sqrt {73}}{4}$
Step 6: $x=\frac{-5 - \sqrt {73}}{4}$ or $x=\frac{-5 + \sqrt {73}}{4}$
Step 7: Therefore, the solution set is {$\frac{-5 - \sqrt {73}}{4},\frac{-5+\sqrt {73}}{4}$}.
