What are the solutions to the equation 12-4x+24x^2=0 ?

The solutions to the equation 12-4x+24x^2=0 are x= 1/12+i(sqrt(71))/(12),x= 1/12-i(sqrt(71))/(12)

Find the zeros of 12-4x+24x^2=0

The zeros of 12-4x+24x^2=0 are x= 1/12+i(sqrt(71))/(12),x= 1/12-i(sqrt(71))/(12)

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12 - 4 x + 24 x^{2} = 0

x = \frac{1}{12} + i \frac{71}{12}, x = \frac{1}{12} - i \frac{71}{12}

Solution steps

12 - 4 x + 24 x^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

24 x^{2} - 4 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 24 \cdot 12}{2 \cdot 24}

Simplify

(- 4)^{2} - 4 \cdot 24 \cdot 12 : 471 i

x_{1, 2} = \frac{- ( - 4 ) \pm 4 71 i}{2 \cdot 24}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 4 71 i}{2 \cdot 24}, x_{2} = \frac{- ( - 4 ) - 4 71 i}{2 \cdot 24}

x = \frac{- ( - 4 ) + 4 71 i}{2 \cdot 24} : \frac{1}{12} + i \frac{71}{12}

x = \frac{- ( - 4 ) - 4 71 i}{2 \cdot 24} : \frac{1}{12} - i \frac{71}{12}

The solutions to the quadratic equation are:

x = \frac{1}{12} + i \frac{71}{12}, x = \frac{1}{12} - i \frac{71}{12}

8^{6}

2 x - 1 = 3

6 \cdot 8

(x - 4)^{2} - 9 = 0

What are the solutions to the equation 12-4x+24x^2=0 ?
The solutions to the equation 12-4x+24x^2=0 are x= 1/12+i(sqrt(71))/(12),x= 1/12-i(sqrt(71))/(12)
Find the zeros of 12-4x+24x^2=0
The zeros of 12-4x+24x^2=0 are x= 1/12+i(sqrt(71))/(12),x= 1/12-i(sqrt(71))/(12)