What are the solutions to the equation r(3r-1)=-3 ?

The solutions to the equation r(3r-1)=-3 are r= 1/6+i(sqrt(35))/6 ,r= 1/6-i(sqrt(35))/6

Find the zeros of r(3r-1)=-3

The zeros of r(3r-1)=-3 are r= 1/6+i(sqrt(35))/6 ,r= 1/6-i(sqrt(35))/6

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r (3 r - 1) = - 3

r = \frac{1}{6} + i \frac{35}{6}, r = \frac{1}{6} - i \frac{35}{6}

Solution steps

r (3 r - 1) = - 3

Expand

r (3 r - 1) : 3 r^{2} - r

3 r^{2} - r = - 3

Move

3

to the left side

3 r^{2} - r + 3 = 0

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 \cdot 3 \cdot 3 : 35 i

r_{1, 2} = \frac{- ( - 1 ) \pm 35 i}{2 \cdot 3}

Separate the solutions

r_{1} = \frac{- ( - 1 ) + 35 i}{2 \cdot 3}, r_{2} = \frac{- ( - 1 ) - 35 i}{2 \cdot 3}

r = \frac{- ( - 1 ) + 35 i}{2 \cdot 3} : \frac{1}{6} + i \frac{35}{6}

r = \frac{- ( - 1 ) - 35 i}{2 \cdot 3} : \frac{1}{6} - i \frac{35}{6}

The solutions to the quadratic equation are:

r = \frac{1}{6} + i \frac{35}{6}, r = \frac{1}{6} - i \frac{35}{6}

184 = - 0.01 x^{2} + 200

- 5.5 x^{2} - x + 24 = 0

110026041.7 x + 74055989.52 x^{2} = 0

\frac{x ^{2}}{5} - \frac{4 x}{5} = 1

2 = - x^{2}

What are the solutions to the equation r(3r-1)=-3 ?
The solutions to the equation r(3r-1)=-3 are r= 1/6+i(sqrt(35))/6 ,r= 1/6-i(sqrt(35))/6
Find the zeros of r(3r-1)=-3
The zeros of r(3r-1)=-3 are r= 1/6+i(sqrt(35))/6 ,r= 1/6-i(sqrt(35))/6