What are the solutions to the equation 3r^2+4r+5=0 ?

The solutions to the equation 3r^2+4r+5=0 are r=-2/3+i(sqrt(11))/3 ,r=-2/3-i(sqrt(11))/3

Find the zeros of 3r^2+4r+5=0

The zeros of 3r^2+4r+5=0 are r=-2/3+i(sqrt(11))/3 ,r=-2/3-i(sqrt(11))/3

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3 r^{2} + 4 r + 5 = 0

r = - \frac{2}{3} + i \frac{11}{3}, r = - \frac{2}{3} - i \frac{11}{3}

Solution steps

3 r^{2} + 4 r + 5 = 0

Solve with the quadratic formula

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

Simplify

4^{2} - 4 \cdot 3 \cdot 5 : 211 i

r_{1, 2} = \frac{- 4 \pm 2 11 i}{2 \cdot 3}

Separate the solutions

r_{1} = \frac{- 4 + 2 11 i}{2 \cdot 3}, r_{2} = \frac{- 4 - 2 11 i}{2 \cdot 3}

r = \frac{- 4 + 2 11 i}{2 \cdot 3} : - \frac{2}{3} + i \frac{11}{3}

r = \frac{- 4 - 2 11 i}{2 \cdot 3} : - \frac{2}{3} - i \frac{11}{3}

The solutions to the quadratic equation are:

r = - \frac{2}{3} + i \frac{11}{3}, r = - \frac{2}{3} - i \frac{11}{3}

0 = 9 x^{2} + 3 x - 2

x^{2} + 8 x + 41

x^{2} + x + \frac{1}{3} = \frac{2}{3}

12 x = 100 x^{2} - 2 \times 9.8 \times 5

- 2 x^{2} - 5 x = 7

What are the solutions to the equation 3r^2+4r+5=0 ?
The solutions to the equation 3r^2+4r+5=0 are r=-2/3+i(sqrt(11))/3 ,r=-2/3-i(sqrt(11))/3
Find the zeros of 3r^2+4r+5=0
The zeros of 3r^2+4r+5=0 are r=-2/3+i(sqrt(11))/3 ,r=-2/3-i(sqrt(11))/3