Answer
{$-2.77,1.27$}
Work Step by Step
Step 1: Comparing $2x^{2}+3x-7=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we obtain:
$a=2$, $b=3$, and $c=-7$
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:
$x=\frac{-(3) \pm \sqrt {(3)^{2}-4(2)(-7)}}{2(2)}$
Step 4: $x=\frac{-3 \pm \sqrt {9+56}}{4}$
Step 5: $x=\frac{-3 \pm \sqrt {65}}{4}$
Step 6: $x=\frac{-3 \pm 8.062}{4}$
Step 7: $x=\frac{-3+8.062}{4}$ or $x=\frac{-3-8.062}{4}$
Step 8: $x=\frac{5.062}{4}$ or $x=\frac{-11.062}{4}$
Step 9: $x=1.266\approx1.27$ or $x=-2.766\approx-2.77$
Step 10: Therefore, the solution set is {$-2.77,1.27$}.
