What are the solutions to the equation 4r^2-r-8=0 ?

The solutions to the equation 4r^2-r-8=0 are r=(1+sqrt(129))/8 ,r=(1-sqrt(129))/8

Find the zeros of 4r^2-r-8=0

The zeros of 4r^2-r-8=0 are r=(1+sqrt(129))/8 ,r=(1-sqrt(129))/8

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4 r^{2} - r - 8 = 0

r = \frac{1 + 129}{8}, r = \frac{1 - 129}{8}

Decimal Notation

r = 1.54472 \dots, r = - 1.29472 \dots

Solution steps

4 r^{2} - r - 8 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 4 ( - 8 )}{2 \cdot 4}

(- 1)^{2} - 4 \cdot 4 (- 8) = 129

r_{1, 2} = \frac{- ( - 1 ) \pm 129}{2 \cdot 4}

Separate the solutions

r_{1} = \frac{- ( - 1 ) + 129}{2 \cdot 4}, r_{2} = \frac{- ( - 1 ) - 129}{2 \cdot 4}

r = \frac{- ( - 1 ) + 129}{2 \cdot 4} : \frac{1 + 129}{8}

r = \frac{- ( - 1 ) - 129}{2 \cdot 4} : \frac{1 - 129}{8}

The solutions to the quadratic equation are:

r = \frac{1 + 129}{8}, r = \frac{1 - 129}{8}

\frac{- 2}{1}

2 (2^{x + y})

2 a^{2} - [a + [2 (a - 1) (a + 3)]] = 26

21 n^{3} - 3 n^{2} - 35 n + 5

- 8 x + 3 = - 6 x + 11

What are the solutions to the equation 4r^2-r-8=0 ?
The solutions to the equation 4r^2-r-8=0 are r=(1+sqrt(129))/8 ,r=(1-sqrt(129))/8
Find the zeros of 4r^2-r-8=0
The zeros of 4r^2-r-8=0 are r=(1+sqrt(129))/8 ,r=(1-sqrt(129))/8