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

The solutions to the equation 3r^2+r-2=0 are r= 2/3 ,r=-1

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

The zeros of 3r^2+r-2=0 are r= 2/3 ,r=-1

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

r = \frac{2}{3}, r = - 1

Solution steps

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

Solve with the quadratic formula

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

1^{2} - 4 \cdot 3 (- 2) = 5

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

Separate the solutions

r_{1} = \frac{- 1 + 5}{2 \cdot 3}, r_{2} = \frac{- 1 - 5}{2 \cdot 3}

r = \frac{- 1 + 5}{2 \cdot 3} : \frac{2}{3}

r = \frac{- 1 - 5}{2 \cdot 3} : - 1

The solutions to the quadratic equation are:

r = \frac{2}{3}, r = - 1

9 v^{4} - 7 v^{2}

p^{2} - 25 = 0

x^{4} + 2 x^{3} + 2 x^{2}

∣ x - 3 ∣ < 3

x^{2} + 3 x y + 2 y^{2}

What are the solutions to the equation 3r^2+r-2=0 ?
The solutions to the equation 3r^2+r-2=0 are r= 2/3 ,r=-1
Find the zeros of 3r^2+r-2=0
The zeros of 3r^2+r-2=0 are r= 2/3 ,r=-1