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Solve the quadratic equation $3 x^{2}+2 x-8=0$
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Given $3 x^{2}+2 x-8=0$,
Note that $\mathrm{a}=3, \mathrm{~b}=2, \mathrm{c}=-8$ and $\mathrm{ac}=-24$
Since $-4 \times 6=-24$ and $-4+6=2$
From $3 x^{2}+2 x-8=0$
Then $\left(3 x^{2}+-4 x\right)(6 x-8)=0$
$x(3 x-4)+2(3 x-4)=0$
$(3 x-4)(x+2)=0$
Either: $3 x-4=0$ or $x+2=0$,
$\mathrm{x}=\frac{44}{33}$ or $\mathrm{x}=-2$.
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