\(2t^2+5t+2\\\\a=2;\ b=5;\ c=2\\\\\Delta=b^2-4ac;\ if\ \Delta > 0\ then\ t_1=\frac{-b-\sqrt\Delta}{2a}\ and\ t_2=\frac{-b+\sqrt\Delta}{2a}\\\\\Delta=5^2-4\cdot2\cdot2=25-16=9;\ \sqrt\Delta=\sqrt9=3\\\\t_1=\frac{-5-3}{2\cdot2}=\frac{-8}{4}=-2;\ t_2=\frac{-5+3}{2\cdot2}=\frac{-2}{4}=-\frac{1}{2}\\\\2t^2+5t+2=2(x+2)(x+\frac{1}{2})\)