Answer
$${\text{ }}x = - \frac{3}{2},\,\,\,\,\,x = 2 \pm \sqrt 3 $$
Work Step by Step
$$\eqalign{
& 2{x^3} - 5{x^2} - 10x + 3 = 0 \cr
& {\text{Let }}p\left( x \right) = 2{x^3} - 5{x^2} - 10x + 3 \cr
& {\text{The constant term is }}3{\text{ }}\left( {{\text{which has divisors }} \pm 1, \pm 3} \right) \cr
& {\text{Evaluating the divisors}} \cr
& p\left( { - 1} \right) = 6 \cr
& p\left( 1 \right) = - 10 \cr
& p\left( { - 3} \right) = - 66 \cr
& p\left( 3 \right) = - 18 \cr
& {\text{The only possible noninteger rational zeros are}}\,\, \pm \frac{2}{3}, \pm \frac{3}{2}, \pm \frac{1}{2}, \pm \frac{1}{3} \cr
& {\text{Evaluating }} \cr
& p\left( { - \frac{2}{3}} \right) = \frac{{185}}{{27}} \cr
& p\left( {\frac{2}{3}} \right) = - \frac{{143}}{{27}} \cr
& p\left( { - \frac{3}{2}} \right) = 0 \cr
& \cr
& p\left( x \right) = 0{\text{ for }}x = - \frac{3}{2},{\text{ then }}\left( {2x + 3} \right){\text{ is a factor of }}p\left( x \right) \cr
& p\left( x \right) = 2{x^3} - 5{x^2} + 10x + 3 = \left( {2x + 3} \right)q\left( x \right) \cr
& {\text{By the long division}}\,\,{\text{ }}2x + 3\left){\vphantom{1{2{x^3} - 5{x^2} + 10x + 3}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{2{x^3} - 5{x^2} + 10x + 3}}}{\text{ we obtain}} \cr
& p\left( x \right) = \left( {2x + 3} \right)\left( {{x^2} - 4x + 1} \right) \cr
& \cr
& Thus, \cr
& 2{x^3} - 5{x^2} - 10x + 3 = 0 \cr
& \left( {2x + 3} \right)\left( {{x^2} - 4x + 1} \right) = 0 \cr
& {\text{By the zero product property}} \cr
& 2x + 3 = 0{\text{ or }}{x^2} - 4x + 1 = 0 \cr
& x = - \frac{3}{2}{\text{ }}\,\,\,\,\,\,{\text{or }}{x^2} - 4x + 1 = 0 \cr
& {\text{By the quadratic formula}} \cr
& {\text{ }}{x^2} - 4x + 1 = 0 \cr
& {\text{ }}x = \frac{{ - \left( { - 4} \right) \pm \sqrt {{{\left( { - 4} \right)}^2} - 4\left( 1 \right)\left( 1 \right)} }}{{2\left( 1 \right)}} \cr
& {\text{ }}x = 2 \pm \sqrt 3 \cr
& \cr
& {\text{then the real solutions are:}} \cr
& {\text{ }}x = - \frac{3}{2},\,\,\,\,\,x = 2 \pm \sqrt 3 \cr} $$
