What are the solutions to the equation 12x^2-3x+2=0 ?

The solutions to the equation 12x^2-3x+2=0 are x= 1/8+i(sqrt(87))/(24),x= 1/8-i(sqrt(87))/(24)

Find the zeros of 12x^2-3x+2=0

The zeros of 12x^2-3x+2=0 are x= 1/8+i(sqrt(87))/(24),x= 1/8-i(sqrt(87))/(24)

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12 x^{2} - 3 x + 2 = 0

x = \frac{1}{8} + i \frac{87}{24}, x = \frac{1}{8} - i \frac{87}{24}

Solution steps

12 x^{2} - 3 x + 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 12 \cdot 2}{2 \cdot 12}

Simplify

(- 3)^{2} - 4 \cdot 12 \cdot 2 : 87 i

x_{1, 2} = \frac{- ( - 3 ) \pm 87 i}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 87 i}{2 \cdot 12}, x_{2} = \frac{- ( - 3 ) - 87 i}{2 \cdot 12}

x = \frac{- ( - 3 ) + 87 i}{2 \cdot 12} : \frac{1}{8} + i \frac{87}{24}

x = \frac{- ( - 3 ) - 87 i}{2 \cdot 12} : \frac{1}{8} - i \frac{87}{24}

The solutions to the quadratic equation are:

x = \frac{1}{8} + i \frac{87}{24}, x = \frac{1}{8} - i \frac{87}{24}

2 c^{2} - 9 c + 5 = 2 c

6 n^{2} + 10 n + 25 = 5 n^{2}

4 y^{2} - 8 y - 10 = - 5

x^{2} + 10 x = - 4 x - 40

\frac{1}{36} - x^{2} = 0

What are the solutions to the equation 12x^2-3x+2=0 ?
The solutions to the equation 12x^2-3x+2=0 are x= 1/8+i(sqrt(87))/(24),x= 1/8-i(sqrt(87))/(24)
Find the zeros of 12x^2-3x+2=0
The zeros of 12x^2-3x+2=0 are x= 1/8+i(sqrt(87))/(24),x= 1/8-i(sqrt(87))/(24)