Answer
$$x = - 7{\text{ and }}x = - 3$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^3} + 15{x^2} + 63x - 10 \cr
& {\text{find the derivative of the function}} \cr
& f'\left( x \right) = {D_x}\left( {{x^3} + 15{x^2} + 63x - 10} \right) \cr
& {\text{use the power rule for derivatives}} \cr
& f'\left( x \right) = 3{x^2} + 15\left( {2x} \right) + 63\left( 1 \right) - 0 \cr
& f'\left( x \right) = 3{x^2} + 30x + 63 \cr
& \cr
& {\text{Find the points of }}x{\text{ where the slope of the tangent line is horizontal}}{\text{, set }}f'\left( x \right) = 0 \cr
& {\text{Then}} \cr
& 3{x^2} + 30x + 63 = 0 \cr
& {\text{factoring}} \cr
& 3\left( {{x^2} + 10x + 21} \right) = 0 \cr
& 3\left( {x + 7} \right)\left( {x + 3} \right) = 0 \cr
& {\text{solving we obtain}} \cr
& x = - 7{\text{ and }}x = - 3 \cr} $$
