\(2v^2+12v-27=9v\ \ \ \ /-9v\\\\2v^2+3v-27=0\\\\a=2;\ b=3;\ c=-27\\\\\Delta=b^2-4ac\\\\\Delta=3^2-4\cdot2\cdot(-27)=9+216=225\\\\v_1=\frac{-b-\sqrt\Delta}{2a};\ v_2=\frac{-b+\sqrt\Delta}{2a}\\\\v_1=\frac{-3-\sqrt{225}}{2\cdot2}=\frac{-3-15}{4}=\frac{-18}{4}=-4\frac{1}{2}\\\\v_2=\frac{-3+15}{2\cdot2}=\frac{12}{4}=3\)
\(2v^2+3v-27=0\\\\2(v+4\frac{1}{2})(v-3)=0\\\\Answer:v=-4\frac{1}{2}\ \vee\ v=3.\)