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

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

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

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

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

r = \frac{- 1 + 2}{2}, r = - \frac{1 + 2}{2}

Decimal Notation

r = 0.20710 \dots, r = - 1.20710 \dots

Solution steps

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

Solve with the quadratic formula

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

4^{2} - 4 \cdot 4 (- 1) = 42

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

r = \frac{- 1 + 2}{2}, r = - \frac{1 + 2}{2}

(x + 7) (x - 3) = 0

13500 x^{2} - 2481 x - 660.75 = 0

5 x (1 - x) = x

k^{2} = k

x^{2} + 4 x + 3 = 3

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