Answer
The solutions are $x=1\pm\dfrac{2\sqrt{6}}{3}$
Work Step by Step
$-3x^{2}+6x+5=0$
Divide the whole equation by $-3$:
$-\dfrac{1}{3}(-3x^{2}+6x+5=0)$
$x^{2}-2x-\dfrac{5}{3}=0$
Take $\dfrac{5}{3}$ to the right side:
$x^{2}-2x=\dfrac{5}{3}$
Complete the square by adding $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation and simplifying. In this case, $b=-2$
$x^{2}-2x+\Big(-\dfrac{2}{2}\Big)^{2}=\dfrac{5}{3}+\Big(-\dfrac{2}{2}\Big)^{2}$
$x^{2}-2x+1=\dfrac{5}{3}+1$
$x^{2}-2x+1=\dfrac{8}{3}$
Factor the left side of the equation, which is a perfect square trinomial:
$(x-1)^{2}=\dfrac{8}{3}$
Take the square root of both sides:
$\sqrt{(x-1)^{2}}=\pm\sqrt{\dfrac{8}{3}}$
$x-1=\pm\dfrac{\sqrt{24}}{3}$
$x-1=\pm\dfrac{2\sqrt{6}}{3}$
Solve for $x$:
$x=1\pm\dfrac{2\sqrt{6}}{3}$
