#### Answer

$$x = \frac{{5 + \sqrt 7 }}{3}{\text{ and }}x = \frac{{5 - \sqrt 7 }}{3}$$

#### Work Step by Step

$$\eqalign{
& f\left( x \right) = {x^3} - 5{x^2} + 6x + 3 \cr
& {\text{find the derivative of the function}} \cr
& f'\left( x \right) = {D_x}\left( {{x^3} - 5{x^2} + 6x + 3} \right) \cr
& {\text{use the power rule for derivatives}} \cr
& f'\left( x \right) = 3{x^2} - 5\left( {2x} \right) + 6\left( 1 \right) + 0 \cr
& f'\left( x \right) = 3{x^2} - 10x + 6 \cr
& \cr
& {\text{Find the points of }}x{\text{ where the slope of the tangent line is horizontal}}{\text{, set }}f'\left( x \right) = 0 \cr
& {\text{Then}} \cr
& 3{x^2} - 10x + 6 = 0 \cr
& {\text{use the quadratic formula }}x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \cr
& x = \frac{{ - \left( { - 10} \right) \pm \sqrt {{{\left( { - 10} \right)}^2} - 4\left( 3 \right)\left( 6 \right)} }}{{2\left( 3 \right)}} \cr
& {\text{simplifying}} \cr
& x = \frac{{10 \pm \sqrt {28} }}{6} \cr
& x = \frac{{10 \pm \sqrt {4 \cdot 7} }}{6} \cr
& x = \frac{{10 \pm 2\sqrt 7 }}{6} \cr
& x = \frac{{5 \pm \sqrt 7 }}{3} \cr
& {\text{then}} \cr
& x = \frac{{5 + \sqrt 7 }}{3}{\text{ and }}x = \frac{{5 - \sqrt 7 }}{3} \cr} $$