Answer
{$\frac{-7 + \sqrt {97}}{4},\frac{-7 - \sqrt {97}}{4}$}
Work Step by Step
Step 1: Comparing $2x^{2}+7x-6=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we obtain:
$a=2$, $b=7$, 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, we obtain:
$x=\frac{-(7) \pm \sqrt {(7)^{2}-4(2)(-6)}}{2(2)}$
Step 4: $x=\frac{-7 \pm \sqrt {49+48}}{4}$
Step 5: $x=\frac{-7 \pm \sqrt {97}}{4}$
Step 6: $x=\frac{-7 + \sqrt {97}}{4}$ or $x=\frac{-7 - \sqrt {97}}{4}$
Step 7: Therefore, the solution set is {$\frac{-7 + \sqrt {97}}{4},\frac{-7 - \sqrt {97}}{4}$}.
