Question

Solve the given equation: $21x^{2} -28x+10=0$

Answer 1

21x² - 28x + 10 = 0

Solve using quadratic formula:

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac} }{2a}$

Sub a = 21, b = -28, c = 10

$x = \dfrac{28 \pm \sqrt{(-28)^2 - 4(21)(10)} }{2(21)}$

Simplify:

$x = \dfrac{28 \pm \sqrt{784 - 840} }{42}$

$x = \dfrac{28 \pm \sqrt{-56} }{42}$

$x = \dfrac{28 \pm 2i\sqrt{14} }{42}$

$x = \dfrac{2}{3} \pm \dfrac{i\sqrt{14} }{21}$

$x = \dfrac{2}{3} + \dfrac{i\sqrt{14} }{21} \ or \ x = \dfrac{2}{3} - \dfrac{i\sqrt{14} }{21}$

Answer 2

Answer:

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