Question
Show the following quadratic equation by factorization method: $21x^2 - 28x + 10 = 0$
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Answer
$21 x^2-28 x+10=0$ We will apply discriminant rule, $x=\frac{-b \pm \sqrt{D}}{2 a} \ldots(A)$
Where $D=b^2-4 a c=(-28)^2-4.21 .10=784$ $-840=-56 \operatorname{From}(A) \mathrm{x}=\frac{-28 \pm \sqrt{-56}}{2.21}=\frac{-28 \pm 2 \sqrt{14} \mathrm{i}}{42}$
$\therefore \mathrm{x}=\frac{2}{3} \pm \frac{\sqrt{14}}{21} \mathrm{i}$
Where $D=b^2-4 a c=(-28)^2-4.21 .10=784$ $-840=-56 \operatorname{From}(A) \mathrm{x}=\frac{-28 \pm \sqrt{-56}}{2.21}=\frac{-28 \pm 2 \sqrt{14} \mathrm{i}}{42}$
$\therefore \mathrm{x}=\frac{2}{3} \pm \frac{\sqrt{14}}{21} \mathrm{i}$
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