What are the solutions to the equation -2r^2+r+7=0 ?

The solutions to the equation -2r^2+r+7=0 are r=-(-1+sqrt(57))/4 ,r=(1+sqrt(57))/4

Find the zeros of -2r^2+r+7=0

The zeros of -2r^2+r+7=0 are r=-(-1+sqrt(57))/4 ,r=(1+sqrt(57))/4

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- 2 r^{2} + r + 7 = 0

r = - \frac{- 1 + 57}{4}, r = \frac{1 + 57}{4}

Decimal Notation

r = - 1.63745 \dots, r = 2.13745 \dots

Solution steps

- 2 r^{2} + r + 7 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 2 ) \cdot 7}{2 ( - 2 )}

1^{2} - 4 (- 2) \cdot 7 = 57

r_{1, 2} = \frac{- 1 \pm 57}{2 ( - 2 )}

Separate the solutions

r_{1} = \frac{- 1 + 57}{2 ( - 2 )}, r_{2} = \frac{- 1 - 57}{2 ( - 2 )}

r = \frac{- 1 + 57}{2 ( - 2 )} : - \frac{- 1 + 57}{4}

r = \frac{- 1 - 57}{2 ( - 2 )} : \frac{1 + 57}{4}

The solutions to the quadratic equation are:

r = - \frac{- 1 + 57}{4}, r = \frac{1 + 57}{4}

x^{2} - 4 x - 60 = 9

k^{3} - r^{3} = (k^{2} + r) (k - r^{2})

4 x^{2} - 4 x = 5

3 t^{2} - 28 t + 49 = 0

x^{2} + (\frac{5 - 2 x}{3})^{2} = 2

What are the solutions to the equation -2r^2+r+7=0 ?
The solutions to the equation -2r^2+r+7=0 are r=-(-1+sqrt(57))/4 ,r=(1+sqrt(57))/4
Find the zeros of -2r^2+r+7=0
The zeros of -2r^2+r+7=0 are r=-(-1+sqrt(57))/4 ,r=(1+sqrt(57))/4