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In equation ${3 x}^{2} + 11 x - 20 = 0$ , value of $x = ?$

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x = \frac{4}{9} or x = - 1

x = \frac{4}{5} or x = - 3

x = \frac{4}{3} or x = - 5

x = \frac{4}{5} or x = - 7

answer is C.

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Detailed Solution

{3 x}^{2} + 11 x - 20 = 0

x = \frac{- 11 \pm \sqrt{11^{2} - 4 \times 3 (- 20)}}{2 \times 3}

[∵ \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a}]

x = \frac{- 11 \pm \sqrt{121 - 4 \times (- 20)}}{2 \times 3}

x = \frac{- 11 \pm \sqrt{121 - 12 (- 20)}}{2 \times 3}

x = \frac{- 11 \pm \sqrt{121 + 240}}{2 \times 3}

x = \frac{- 11 \pm \sqrt{361}}{2 \times 3}

x = \frac{- 11 \pm 19}{2 \times 3}

x = \frac{- 11 \pm 19}{6}

Now when we solve the equation

x = \frac{- 11 \pm 19}{6}

, when

\pm

is plus. Add

- 11 to 19

x = \frac{8}{6} = \frac{4}{3}

Now when we solve the equation

x = \frac{- 11 \pm 19}{6}

, when

\pm

is minus. Subtract

19

from

- 11

x = \frac{- 30}{6} = - 5

So,

x = \frac{4}{3} or x = - 5

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