What are the solutions to the equation 3c^2+4c+8=6c ?

The solutions to the equation 3c^2+4c+8=6c are c= 1/3+i(sqrt(23))/3 ,c= 1/3-i(sqrt(23))/3

Find the zeros of 3c^2+4c+8=6c

The zeros of 3c^2+4c+8=6c are c= 1/3+i(sqrt(23))/3 ,c= 1/3-i(sqrt(23))/3

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3 c^{2} + 4 c + 8 = 6 c

c = \frac{1}{3} + i \frac{23}{3}, c = \frac{1}{3} - i \frac{23}{3}

Solution steps

3 c^{2} + 4 c + 8 = 6 c

Move

6 c

to the left side

3 c^{2} - 2 c + 8 = 0

Solve with the quadratic formula

c_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 3 \cdot 8}{2 \cdot 3}

Simplify

(- 2)^{2} - 4 \cdot 3 \cdot 8 : 223 i

c_{1, 2} = \frac{- ( - 2 ) \pm 2 23 i}{2 \cdot 3}

Separate the solutions

c_{1} = \frac{- ( - 2 ) + 2 23 i}{2 \cdot 3}, c_{2} = \frac{- ( - 2 ) - 2 23 i}{2 \cdot 3}

c = \frac{- ( - 2 ) + 2 23 i}{2 \cdot 3} : \frac{1}{3} + i \frac{23}{3}

c = \frac{- ( - 2 ) - 2 23 i}{2 \cdot 3} : \frac{1}{3} - i \frac{23}{3}

The solutions to the quadratic equation are:

c = \frac{1}{3} + i \frac{23}{3}, c = \frac{1}{3} - i \frac{23}{3}

f = 12 (2)^{\frac{- 3}{500}}

8 x^{2} + 5 x + \frac{13}{4} = 0

(X + 7)^{2} + 12 = 0

(\frac{2}{2})^{2}

6 x^{2} - 14 x + 8

What are the solutions to the equation 3c^2+4c+8=6c ?
The solutions to the equation 3c^2+4c+8=6c are c= 1/3+i(sqrt(23))/3 ,c= 1/3-i(sqrt(23))/3
Find the zeros of 3c^2+4c+8=6c
The zeros of 3c^2+4c+8=6c are c= 1/3+i(sqrt(23))/3 ,c= 1/3-i(sqrt(23))/3