Finding roots of $2x^3-5x^2+18x+45$

Question

solve $2x^3-5x^2+18x+45$
not exactly sure where to start on finding the zeros complex or real. There is one real zero and two complex I know that from graphing just cannot do it on paper to understand it.

Answer 1

You can remark that $\frac{-3}{2}=-1.5$ is a solution.

Answer 2

From Hamza's post one solution is -1.5,others you can find dividing polynomial to the $(x+1.5)$,then we get: $$2x^{ 3 }-5x^{ 2 }+18x+45=0\\ \left( x+1.5 \right) \left( 2x^{ 2 }-8x+30 \right) =0\\ \left( x+1.5 \right) \left( x^{ 2 }-4x+15 \right) =0\\ { x }_{ 1 }=-1.5,x_{ 2 }=2+i\sqrt { 11 } ,{ x }_{ 3 }=2-i\sqrt { 11 } $$