What are the real roots of 2x^3+9x^2+4x-15=0

Well, luckily it is apparent that (x-1) is a root because when x=1 the equation is equal to zero.  So we can divide the equation by that factor to find the other roots. 

(2x^3+9x^2+4x-15)/(x-1)
2x^2  r  11x^2+4x-15
11x    r  15x-15
15     r   0

(x-1)(2x^2+11x+15)=0

(x-1)(2x^2+6x+5x+15)=0

(x-1)(2x(x+3)+5(x+3))=0

(x-1)(2x+5)(x+3)=0

So the roots are x= -3, -2.5, 1