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 = 1.68614 \dots, r = - 1.18614 \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}

x, 12 x^{2} - 288 x + 1296 = 0

3 (x - 5)^{2} - 12 = 0

3 x^{2} + 2 x - 1 = 0

(x - 5) (2 - x) = 0

5 x^{2} - 12 x = 9

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