Answer
{$-\frac{5}{3},1$}
Work Step by Step
Step 1: $3x^{2}+2x=5$
Step 2: Subtracting 5 from both sides, $3x^{2}+2x-5=5-5$
Step 3: The equation therefore becomes $3x^{2}+2x-5=0$
Step 4: Comparing $3x^{2}+2x-5=0$ to the standard form of a quadratic equation $ax^{2}+bx+c=0$;
$a=3$, $b=2$ and $c=-5$
Step 5: The quadratic formula is:
$x=\frac{-b \pm \sqrt (b^{2}-4ac)}{2a}$
Step 6: Substituting the values of a,b and c in the formula:
$x=\frac{-(2) \pm \sqrt ((2)^{2}-4(3)(-5))}{2(3)}$
Step 7: $x=\frac{-2 \pm \sqrt (4+60)}{6}$
Step 8: $x=\frac{-2 \pm \sqrt (64)}{6}$
Step 9: $x=\frac{-2 \pm 8}{6}$
Step 10: $x=\frac{-2+8}{6}$ or $x=\frac{-2-8}{6}$
Step 11: $x=\frac{6}{6}$ or $x=\frac{-10}{6}$
Step 12: $x=1$ or $x=\frac{-5}{3}$
Step 13: Therefore, the solution set is {$-\frac{5}{3},1$}.
