What are the solutions to the equation -12r^2-2=-4r ?

The solutions to the equation -12r^2-2=-4r are r= 1/6-i(sqrt(5))/6 ,r= 1/6+i(sqrt(5))/6

Find the zeros of -12r^2-2=-4r

The zeros of -12r^2-2=-4r are r= 1/6-i(sqrt(5))/6 ,r= 1/6+i(sqrt(5))/6

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- 12 r^{2} - 2 = - 4 r

r = \frac{1}{6} - i \frac{5}{6}, r = \frac{1}{6} + i \frac{5}{6}

Solution steps

- 12 r^{2} - 2 = - 4 r

Move

4 r

to the left side

- 12 r^{2} - 2 + 4 r = 0

Write in the standard form

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

- 12 r^{2} + 4 r - 2 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 ( - 12 ) ( - 2 )}{2 ( - 12 )}

Simplify

4^{2} - 4 (- 12) (- 2) : 45 i

r_{1, 2} = \frac{- 4 \pm 4 5 i}{2 ( - 12 )}

Separate the solutions

r_{1} = \frac{- 4 + 4 5 i}{2 ( - 12 )}, r_{2} = \frac{- 4 - 4 5 i}{2 ( - 12 )}

r = \frac{- 4 + 4 5 i}{2 ( - 12 )} : \frac{1}{6} - i \frac{5}{6}

r = \frac{- 4 - 4 5 i}{2 ( - 12 )} : \frac{1}{6} + i \frac{5}{6}

The solutions to the quadratic equation are:

r = \frac{1}{6} - i \frac{5}{6}, r = \frac{1}{6} + i \frac{5}{6}

0.825 x^{2} + 0.15 x - 0.003621753 = 0

x, f (x + c) = 8 (x + c)^{2} + 4

49 k^{2} - 231 k - 70 = 0

a x^{2} + a x + 7 = 3 x^{2} - x + 5 a

mn x^{2} - (m^{2} - n^{2}) x + mn = 0

What are the solutions to the equation -12r^2-2=-4r ?
The solutions to the equation -12r^2-2=-4r are r= 1/6-i(sqrt(5))/6 ,r= 1/6+i(sqrt(5))/6
Find the zeros of -12r^2-2=-4r
The zeros of -12r^2-2=-4r are r= 1/6-i(sqrt(5))/6 ,r= 1/6+i(sqrt(5))/6