How do you solve $- 6 x^{2} - 9 x + 2 = 0$ $-6x^2-9x+2=0$ using the quadratic formula?

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Answer 1 · 2018-05-03T14:09:51

x=-1.696484724
or
x=0.1964847243

$x = \frac{- (- 9) \pm \sqrt{9^{2} - (4 \times (- 6) \times 2)}}{2 \times (- 6)}$ $x={-(-9)\pmsqrt[9^2-(4xx(-6)xx2)]}/(2xx(-6))$

$x = \frac{9 \pm \sqrt{81 + 48}}{- 12}$ $x={9\pmsqrt[81+48]}/(-12)$

$x = \frac{9 \pm \sqrt{129}}{- 12}$ $x={9\pmsqrt[129]}/(-12)$

$x = \frac{9 + \sqrt{129}}{- 12}$ $x={9+sqrt[129]}/(-12)$ $\Rightarrow x = - 1.696484724$ $=> x=-1.696484724$

or

$x = \frac{9 - \sqrt{129}}{- 12}$ $x={9-sqrt[129]}/(-12)$ $\Rightarrow x = 0.1964847243$ $=> x=0.1964847243$

How do you solve -6x^2-9x+2=0−6x2−9x+2=0-6x^2-9x+2=0 using the quadratic formula?