How do you solve using the quadratic formula $3 x^{2} + 4 x - 2 = 0$ $3x^2 + 4x - 2 = 0$ ?

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Answer 1 · 2015-05-16T07:55:15

x= $\frac{- 4 \pm \sqrt{16 + 24}}{6}$ $(-4+-sqrt(16 + 24))/6$

= $\frac{- 4 \pm 2 \sqrt{10}}{6}$ $(-4+-2sqrt10)/6$

= $\frac{- 2 \pm \sqrt{10}}{3}$ $(-2+-sqrt10)/3$

How do you solve using the quadratic formula 3x^2 + 4x - 2 = 0 3x2+4x−2=03x^2 + 4x - 2 = 0 ?