Solve the equation $$3x^2-4x+\dfrac{20}{3}=0$$

Asked by Pragya Singh | 1 year ago |  67

##### Solution :-

Given quadratic equation, $$3x^2 – 4x + \dfrac{20}{3} = 0$$

It can be re-written as: 9x2 – 12x + 20 = 0

On comparing it with ax2 + bx + c = 0, we get

a = 9, b = –12, and c = 20

So, the discriminant of the given equation will be

D = b2 – 4ac = (–12)2 – 4 × 9 × 20

= 144 – 720 = –576

Hence, the required solutions are

$$\dfrac{-b\pm\sqrt{D}}{2a}$$

$$\dfrac{-(12)\pm\sqrt{-576}}{2\times 9}$$

$$\dfrac{12\pm\sqrt{576i}}{18}$$

$$\dfrac{12\pm 24i}{18}$$

$$\dfrac{6(2\pm 4i)}{18}$$

$$\dfrac{2\pm 4i}{3}$$

$$\dfrac{2}{3}\pm \dfrac{4}{3}i$$

Hence, solved

Answered by Abhisek | 1 year ago

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