Solve:
$2 x^{2} + 6 x + 3 = 0, - 3 x^{2} - x - 4 = 0, 1 + 2 x - 3 x^{2} = 0.$

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Solution

$(i) 2 x^{2} + 6 x + 3 = 0$
finding $b^{2} - 4 a c$
$36 - 4 (2) (3)$
$12$
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
$= \frac{- 6 \pm \sqrt{12}}{4} = \frac{- 3 \pm \sqrt{3}}{2}$
$= \frac{- 3 + \sqrt{3}}{2}, \frac{- 3 - \sqrt{3}}{2}$

$(i i) - 3 x^{2} - x - 4 = 0$
Finding $b^{2} - 4 a c$
$(1)^{2} - 4 (- 3) (- 4)$
$1 - 48 = - 47$
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
$= \frac{- (- 1) \pm \sqrt{47}}{2 (- 3)}$
$= \frac{1 \pm \sqrt{47}!}{- 6} = \frac{(1 + \sqrt{47}!)}{6}, \frac{- (1 - \sqrt{47}!)}{6}$

$(i i i) - 3 x^{2} + 2 x + 1$
Finding $b^{2} - 4 a c$
$(2)^{2} - 4 (- 3) (1) = 4 + 12 = 16$
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
$= \frac{- 2 \pm \sqrt{16}}{2 (- 3)} = \frac{- 2 \pm 4}{- 6} = \frac{- 6}{- 6}, \frac{2}{- 6} = + 1, \frac{- 1}{3}$

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Solve:2x2+6x+3=0,−3x2−x−4=0,1+2x−3x2=0.

Solve:
$2 x^{2} + 6 x + 3 = 0, - 3 x^{2} - x - 4 = 0, 1 + 2 x - 3 x^{2} = 0.$