Therefore either [tex]x = \frac{-3+\sqrt{17}}{2}[/tex] or, [tex]x = \frac{-3-\sqrt{17}}{2}[/tex]
Step-by-step explanation:
[tex]3x^2 +9x -6=0[/tex] [tex]x = \frac{-b\pm\sqrt{b^2- 4ac} }{2a}[/tex]
[tex]\Leftrightarrow x = \frac{-9\pm\sqrt{9^2- 4.3(-6)} }{2.3}[/tex] here a = 3 ,b = 9 and c= -6
[tex]\Leftrightarrow x = \frac{-9\pm\sqrt{153} }{6}[/tex]
[tex]\Leftrightarrow x = \frac{3(-3\pm\sqrt{17}) }{6}[/tex]
[tex]\Leftrightarrow x = \frac{-3\pm\sqrt{17}}{2}[/tex]
Therefore either [tex]x = \frac{-3+\sqrt{17}}{2}[/tex] or, [tex]x = \frac{-3-\sqrt{17}}{2}[/tex]
