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

The solutions to the equation 2r^2+r-2=0 are r=(-1+sqrt(17))/4 ,r=(-1-sqrt(17))/4

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

The zeros of 2r^2+r-2=0 are r=(-1+sqrt(17))/4 ,r=(-1-sqrt(17))/4

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

r = \frac{- 1 + 17}{4}, r = \frac{- 1 - 17}{4}

Decimal Notation

r = 0.78077 \dots, r = - 1.28077 \dots

Solution steps

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

Solve with the quadratic formula

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

1^{2} - 4 \cdot 2 (- 2) = 17

r_{1, 2} = \frac{- 1 \pm 17}{2 \cdot 2}

Separate the solutions

r_{1} = \frac{- 1 + 17}{2 \cdot 2}, r_{2} = \frac{- 1 - 17}{2 \cdot 2}

r = \frac{- 1 + 17}{2 \cdot 2} : \frac{- 1 + 17}{4}

r = \frac{- 1 - 17}{2 \cdot 2} : \frac{- 1 - 17}{4}

The solutions to the quadratic equation are:

r = \frac{- 1 + 17}{4}, r = \frac{- 1 - 17}{4}

y, x^{2} + 2 y^{2} = 6

x^{2} = \frac{1}{4} \frac{4 3 + 1}{4}

m^{2} + m - 9 = 0

15 + (0.5) (1.06) (10)^{2} = 30 + (0.5) (1.06) v^{2}

x, x^{2} + y^{2} + z^{2} = c^{2}

What are the solutions to the equation 2r^2+r-2=0 ?
The solutions to the equation 2r^2+r-2=0 are r=(-1+sqrt(17))/4 ,r=(-1-sqrt(17))/4
Find the zeros of 2r^2+r-2=0
The zeros of 2r^2+r-2=0 are r=(-1+sqrt(17))/4 ,r=(-1-sqrt(17))/4