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Solve for $x$ in $12 x^{3}-4 x^{2}-5 x=0$
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$12 x^{3}-4 x^{2}-5 x=0$

start by factorizing with the common factor $x$

$x\left(12 x^{2}-4 x-5\right)=0$

Using the rule: If $a\cdot b =0$ then $a=0$ or $b=0$ or both are equal to zero.

then,

$\Rightarrow x=0$

or

$12 x^{2}-4 x-5=0$

$x=\dfrac{\pm \sqrt{b^{2}-4 a c}}{2a}$

$a=12; b=-4; c=-5$

$x= \dfrac{-(-4) \pm \sqrt{(-4)^{2}-4(12)(-5)}}{2(12)}$

$= \dfrac{4 \pm \sqrt{256}}{24}$

$= \dfrac{4 \pm 16}{24}$

$= \dfrac{ 4 \pm 2}{3}$

$\therefore x=0 \quad \text{OR} \quad x=\dfrac{4+2}{3} \quad \text{OR} \quad x=\dfrac{4-2}{3}$
by Platinum (131,378 points)
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