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+sqrt(19))/3 ,r=(2-sqrt(19))/3

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

The zeros of 3r^2-4r-5=0 are r=(2+sqrt(19))/3 ,r=(2-sqrt(19))/3

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

r = \frac{2 + 19}{3}, r = \frac{2 - 19}{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 ( - 5 )}{2 \cdot 3}

(- 4)^{2} - 4 \cdot 3 (- 5) = 219

r_{1, 2} = \frac{- ( - 4 ) \pm 2 19}{2 \cdot 3}

Separate the solutions

r_{1} = \frac{- ( - 4 ) + 2 19}{2 \cdot 3}, r_{2} = \frac{- ( - 4 ) - 2 19}{2 \cdot 3}

r = \frac{- ( - 4 ) + 2 19}{2 \cdot 3} : \frac{2 + 19}{3}

r = \frac{- ( - 4 ) - 2 19}{2 \cdot 3} : \frac{2 - 19}{3}

The solutions to the quadratic equation are:

r = \frac{2 + 19}{3}, r = \frac{2 - 19}{3}

3 x^{2} + 7 x - 3 = 0

x^{2} + 40 = 4

100 x^{2} - 25 = 0

(x - 7)^{2} = 32

- 2 x^{2} - 6 x + 8 = 0

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+sqrt(19))/3 ,r=(2-sqrt(19))/3
Find the zeros of 3r^2-4r-5=0
The zeros of 3r^2-4r-5=0 are r=(2+sqrt(19))/3 ,r=(2-sqrt(19))/3