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Using the factoring method, solve the quadratic equation: $2 x^{2}-3 x=0$
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$2 \mathrm{x}^{2}-3 \mathrm{x}=0 \ldots$ we see that both of the terms contain $\mathrm{x}$; so we can take it out as a factor:
$x(2 x-3)=0 \ldots$ two terms are multiplied and the result is zero. This means that either of the terms or both of the
terms can be equal to zero:
$\mathrm{x}=0 \ldots$ this is one solution
$2 x-3=0$
$2 x=3$
$x=3 / 2$
$x=1.5 \ldots$ this is the second solution.
So, the solutions are 0 and $1.5 .$
ago by Diamond (78,880 points)

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