Answer
{$\frac{3 \pm 2\sqrt (6)}{3}$}
Work Step by Step
Step 1: Comparing $-3x^{2}+6x+5=0$ to the standard form of a quadratic equation $ax^{2}+bx+c=0$;
$a=-3$, $b=6$ and $c=5$
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:
$x=\frac{-(6) \pm \sqrt ((6)^{2}-4(-3)(5))}{2(-3)}$
Step 4: $x=\frac{-6 \pm \sqrt (36+60)}{-6}$
Step 5: $x=\frac{-6 \pm \sqrt (96)}{-6}$
Step 6: $x=\frac{-6 \pm \sqrt (16\times6)}{-6}$
Step 7: $x=\frac{-6 \pm 4\sqrt (6)}{-6}$
Step 8: $x=\frac{2(-3 \pm 2\sqrt (6))}{-6}$
Step 9: $x=\frac{-3 \pm 2\sqrt (6)}{-3}$
Step 10: $x=\frac{-3+2\sqrt (6)}{-3}$ or $x=\frac{-3-2\sqrt (6)}{-3}$
Step 11: $x=\frac{-1(3-2\sqrt (6))}{-3}$ or $x=\frac{-1(3+2\sqrt (6))}{-3}$
Step 12: $x=\frac{3-2\sqrt (6)}{3}$ or $x=\frac{3+2\sqrt (6)}{3}$
Step 13: Therefore, the solution set is {$\frac{3 \pm 2\sqrt (6)}{3}$}.
