Finding the real root of the polynomial $2x^3-3x^2+2 $ I want to get exactly roots of this equation... 
$2x^3-3x^2+2 = 0$
I try to solve it but can not find the solution.
wolframealpha just give me aproximation.. 
I know the real root is $-1< root <-1/2$.

Here are the three exact solutions, which you can find by cubic root methods such as Cardano's:
$\left\{\frac{1}{2} \left(1-\frac{1}{\sqrt[3]{3-2 \sqrt{2}}}-\sqrt[3]{3-2
   \sqrt{2}}\right),\frac{1}{2}+\frac{1}{4} \sqrt[3]{3-2 \sqrt{2}} \left(1-i
   \sqrt{3}\right)+\frac{1+i \sqrt{3}}{4 \sqrt[3]{3-2 \sqrt{2}}},\frac{1}{2}+\frac{1-i
   \sqrt{3}}{4 \sqrt[3]{3-2 \sqrt{2}}}+\frac{1}{4} \sqrt[3]{3-2 \sqrt{2}} \left(1+i
   \sqrt{3}\right)\right\}$ 
The first root is real, and has a decimal approximation of $-0.677651...$.
