Math, asked by PragyaTbia, 1 year ago

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

Answers

Answered by TooFree
4

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}

Answered by manitasweetylakra
0

Answer:

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