Solve 21x^2- 28x + 10 = 0 - Vidyadip

Question

Solve $21x^2- 28x + 10 = 0$

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Answer

$21x^2 - 28x + 10 = 0$
Comparing the given quadratic equation with $ax^2 + bx + c = 0$ we have,
$a= 21, b = -28$ and $c = 10$
$\therefore x = \frac{{ - ( - 28) \pm \sqrt {{{( - 28)}^2} - 4 \times 21 \times 10} }}{{2 \times 21}}$
$ = \frac{{28 \pm \sqrt {784 - 840} }}{{42}}$
$ = \frac{{28 \pm \sqrt { - 56} }}{{42}} = \frac{{28 \pm 2\sqrt {14} i}}{{42}}$$ = \frac{{14 \pm \sqrt {14} i}}{{21}}$
Thus $x = \frac{2}{3} + \frac{{\sqrt {14} }}{{21}}i$ and $x = \frac{2}{3} - \frac{{\sqrt {14} }}{{21}}i$

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