What are the solutions to the equation -3r^2-2r-5=0 ?

The solutions to the equation -3r^2-2r-5=0 are r=-1/3-i(sqrt(14))/3 ,r=-1/3+i(sqrt(14))/3

Find the zeros of -3r^2-2r-5=0

The zeros of -3r^2-2r-5=0 are r=-1/3-i(sqrt(14))/3 ,r=-1/3+i(sqrt(14))/3

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- 3 r^{2} - 2 r - 5 = 0

r = - \frac{1}{3} - i \frac{14}{3}, r = - \frac{1}{3} + i \frac{14}{3}

Solution steps

- 3 r^{2} - 2 r - 5 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 3 ) ( - 5 )}{2 ( - 3 )}

Simplify

(- 2)^{2} - 4 (- 3) (- 5) : 214 i

r_{1, 2} = \frac{- ( - 2 ) \pm 2 14 i}{2 ( - 3 )}

Separate the solutions

r_{1} = \frac{- ( - 2 ) + 2 14 i}{2 ( - 3 )}, r_{2} = \frac{- ( - 2 ) - 2 14 i}{2 ( - 3 )}

r = \frac{- ( - 2 ) + 2 14 i}{2 ( - 3 )} : - \frac{1}{3} - i \frac{14}{3}

r = \frac{- ( - 2 ) - 2 14 i}{2 ( - 3 )} : - \frac{1}{3} + i \frac{14}{3}

The solutions to the quadratic equation are:

r = - \frac{1}{3} - i \frac{14}{3}, r = - \frac{1}{3} + i \frac{14}{3}

6 x y^{0}

4 x^{2} + 3 x y - 10 y^{2}

- 7 + 2 x \leq 5

\frac{15 x ^{2} - 23 x - 28}{5 x ^{2} - 6 x - 8} < 0

2 (\frac{x}{2} + \frac{3}{2}) = 2.8

What are the solutions to the equation -3r^2-2r-5=0 ?
The solutions to the equation -3r^2-2r-5=0 are r=-1/3-i(sqrt(14))/3 ,r=-1/3+i(sqrt(14))/3
Find the zeros of -3r^2-2r-5=0
The zeros of -3r^2-2r-5=0 are r=-1/3-i(sqrt(14))/3 ,r=-1/3+i(sqrt(14))/3