What are the solutions to the equation 8r^2+4r-1=0 ?

The solutions to the equation 8r^2+4r-1=0 are r=(-1+sqrt(3))/4 ,r=-(1+sqrt(3))/4

Find the zeros of 8r^2+4r-1=0

The zeros of 8r^2+4r-1=0 are r=(-1+sqrt(3))/4 ,r=-(1+sqrt(3))/4

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8 r^{2} + 4 r - 1 = 0

r = \frac{- 1 + 3}{4}, r = - \frac{1 + 3}{4}

Decimal Notation

r = 0.18301 \dots, r = - 0.68301 \dots

Solution steps

8 r^{2} + 4 r - 1 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 8 (- 1) = 43

r_{1, 2} = \frac{- 4 \pm 4 3}{2 \cdot 8}

Separate the solutions

r_{1} = \frac{- 4 + 4 3}{2 \cdot 8}, r_{2} = \frac{- 4 - 4 3}{2 \cdot 8}

r = \frac{- 4 + 4 3}{2 \cdot 8} : \frac{- 1 + 3}{4}

r = \frac{- 4 - 4 3}{2 \cdot 8} : - \frac{1 + 3}{4}

The solutions to the quadratic equation are:

r = \frac{- 1 + 3}{4}, r = - \frac{1 + 3}{4}

x, x^{2} y - 3 y + 2 = 0

\frac{3}{2} x^{2} + 11 = 59

(5 x + 24) (4 x + 37) = 0

(x + 5) (x + 1) = x - 1

x, f = I n (x^{2} + 7)

What are the solutions to the equation 8r^2+4r-1=0 ?
The solutions to the equation 8r^2+4r-1=0 are r=(-1+sqrt(3))/4 ,r=-(1+sqrt(3))/4
Find the zeros of 8r^2+4r-1=0
The zeros of 8r^2+4r-1=0 are r=(-1+sqrt(3))/4 ,r=-(1+sqrt(3))/4