Social Sciences, asked by rakeshbhaidudwala, 11 months ago

$3 {x }^{2} + 4x - 4 = 0 \\ \\ find \: the \: vaules \: of \: x$

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Answered by abhinay365

1

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Given

$3 {x}^{2} + 4x - 4 = 0 \\ 3 {x}^{2} + 6x - 2x - 4 = 0 \\ 3x(x + 2) - 2(x + 2) = 0 \\ (x + 2)(3x - 2) = 0 \\ x + 2 = 0 \: or \: 3x - 2 = 0 \\ x = - 2 \: or \: 3x = 2 \\ x = - 2 \: or \: x = \frac{2}{3}$

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Answered by ishwarsinghdhaliwal

1

$3x ^{2} + 4x - 4 = 0 \\ 3 {x}^{2} + 6x - 2x - 4 = 0 \\ 3x(x + 2) - 2( x + 2) = 0 \\ (3x - 2)(x + 2) = 0 \\ 3x - 2 = 0 \: \: \: and \: \: \: x + 2 = 0 \\ 3x = 2 \: \: and \: x = - 2 \\ x = \frac{2}{3} \: \: \: \: \: \: \:and \: \: \: \: \: - 2$

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