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

The solutions to the equation 2r^2+r-4=0 are r=(-1+sqrt(33))/4 ,r=(-1-sqrt(33))/4

Find the zeros of 2r^2+r-4=0

The zeros of 2r^2+r-4=0 are r=(-1+sqrt(33))/4 ,r=(-1-sqrt(33))/4

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2 r^{2} + r - 4 = 0

r = \frac{- 1 + 33}{4}, r = \frac{- 1 - 33}{4}

Decimal Notation

r = 1.18614 \dots, r = - 1.68614 \dots

Solution steps

2 r^{2} + r - 4 = 0

Solve with the quadratic formula

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

1^{2} - 4 \cdot 2 (- 4) = 33

r_{1, 2} = \frac{- 1 \pm 33}{2 \cdot 2}

Separate the solutions

r_{1} = \frac{- 1 + 33}{2 \cdot 2}, r_{2} = \frac{- 1 - 33}{2 \cdot 2}

r = \frac{- 1 + 33}{2 \cdot 2} : \frac{- 1 + 33}{4}

r = \frac{- 1 - 33}{2 \cdot 2} : \frac{- 1 - 33}{4}

The solutions to the quadratic equation are:

r = \frac{- 1 + 33}{4}, r = \frac{- 1 - 33}{4}

m^{2} + 5 m - 14 = 0

∣ x - 5 ∣ > 7

2 (2 y + 9) = 50

7 r^{2} + 5 r + 9 = 0

What are the solutions to the equation 2r^2+r-4=0 ?
The solutions to the equation 2r^2+r-4=0 are r=(-1+sqrt(33))/4 ,r=(-1-sqrt(33))/4
Find the zeros of 2r^2+r-4=0
The zeros of 2r^2+r-4=0 are r=(-1+sqrt(33))/4 ,r=(-1-sqrt(33))/4