Question

Solve This Using The Quadratic Formula: 3x^2+9x-6=0

Answer 1

Therefore either  $x = \frac{-3+\sqrt{17}}{2}$    or,  $x = \frac{-3-\sqrt{17}}{2}$

Step-by-step explanation:

$3x^2 +9x -6=0$                                $x = \frac{-b\pm\sqrt{b^2- 4ac} }{2a}$

$\Leftrightarrow x = \frac{-9\pm\sqrt{9^2- 4.3(-6)} }{2.3}$                     here a = 3 ,b = 9 and c= -6

$\Leftrightarrow x = \frac{-9\pm\sqrt{153} }{6}$

$\Leftrightarrow x = \frac{3(-3\pm\sqrt{17}) }{6}$

$\Leftrightarrow x = \frac{-3\pm\sqrt{17}}{2}$

Therefore either  $x = \frac{-3+\sqrt{17}}{2}$    or,  $x = \frac{-3-\sqrt{17}}{2}$