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Solve the equation 21x^2-28x+10=0...

Solve the equation `21x^2-28x+10=0`

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b^(2)-4 a c=(-28)^(2)-4 xx 21 xx 10=784-840=-56.
.therefore. The solutions are given by .cdot x=(28 pm sqrt(-56))/(2 xx 21)=(28 pm sqrt(56) i)/(2 xx 21 |)=(2 xx 14 pm 2 xx sqrt(14))/(2 xx 21).
.=(14)/(21) pm (sqrt(14))/(21)i=(2)/(3) pm (sqrt(14))/(21)i.
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V PUBLICATION-COMPLEX NUMBERS AND QUADRATIC EQUATIONS-QUESTION BANK
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  2. Solve 27 x^(2)-10 x+1=0

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  3. Solve the equation 21x^2-28x+10=0

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