How do you solve $3 x^{2} = 2 (x - 1)$ $3x ^ { 2} = 2( x - 1)$ ?

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How do you solve $3 x^{2} = 2 (x - 1)$ $3x ^ { 2} = 2( x - 1)$ ?

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Answer 1 · 2018-04-22T00:01:33

$x 1 = (\frac{2 + \sqrt{- 20}}{6}) and x 2 = (\frac{2 - \sqrt{- 20}}{6}))$ $x1 = ((2+sqrt(-20))/6) and x2 =((2-sqrt(-20))/6))$

Explanation:

Given:
$3 x^{2} = 2 (x - 1)$ $3x^2 = 2(x-1)$
$3 x^{2} = 2 x - 2$ $3x^2 = 2x - 2$

Rearranging this gives us a Quadratic Equation, that is;
$3 x^{2} - 2 x + 2 = 0$ $3x^2 -2x + 2 = 0$
$x = \frac{- b \pm \sqrt{b^{4} - 4 a c}}{2 a}$ $x = (-b+-sqrt(b^4-4ac))/(2a)$
$x 1 = \frac{- (- 2) + \sqrt{{(- 2)}^{2} - 4 (3) (2)}}{2 \cdot 3}$ $x1 = (-(-2)+sqrt((-2)^2-4(3)(2)))/(2*3)$
$x 2 = \frac{- (- 2) - \sqrt{{(- 2)}^{2} - 4 (3) (2)}}{2 \cdot 3}$ $x2 = (-(-2) -sqrt((-2)^2-4(3)(2)))/(2*3)$
Therefore, $x 1 = \frac{2 + \sqrt{- 20}}{6} and x 2 = \frac{2 - \sqrt{- 20}}{6})$ $x1 = color(red)((2+sqrt(-20))/6) and x2 = color(red)((2-sqrt(-20))/6))$

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How do you solve $3 x^{2} = 2 (x - 1)$ $3x ^ { 2} = 2( x - 1)$ ?

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How do you solve 3x2=2(x−1)3x ^ { 2} = 2( x - 1)?

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How do you solve $3 x^{2} = 2 (x - 1)$ $3x ^ { 2} = 2( x - 1)$ ?