Question

3(4x+8)=2(9-3x) x=?​

Answer 1

$3(4x + 8) = 2(9 - 3x)x$

$12x + 24 = 18x - 6x^{2}$

$18x - 6x^{2} - 12x - 24 = 0$

$6x - 6x^{2} - 24 = 0$

$- (6x^{2} - 6x + 24) = 0$

$6x^{2} - 6x + 24$

$6(x^{2} - x + 4) = 0$

$x^{2} - x + 4 = 0$

$let \: a = 1 \:, b = - 1 \: ,\: c = 4$

$Now, \\ D = b^{2} - 4ac$

$D = ( - 1)^{2} - 4 \times 1 \times 4$

$D = 1 -16$

$D = - 15$

$\therefore \: root \: of \: quadratic \: eqn \: are \: not \: \\ real$

$NOW, \: x = \frac{ - b \frac{ + }{} \sqrt{D} }{2a}$

$x = - \frac{ - ( - 1) \frac{ + }{} \sqrt{ - 15} } {2 \times 1}$

$x = \frac{1 \frac{ + }{ } \sqrt{ - 15} }{2}$

$either , \: x = \frac{1 + \sqrt{ - 15} }{2} \: or \: \\ x = \frac{1 - \sqrt{ - 15} }{2}$

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