What are the solutions to the equation 16r^2+8r+1=0 ?

The solutions to the equation 16r^2+8r+1=0 are r=-1/4

Find the zeros of 16r^2+8r+1=0

The zeros of 16r^2+8r+1=0 are r=-1/4

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16 r^{2} + 8 r + 1 = 0

r = - \frac{1}{4}

Decimal Notation

r = - 0.25

Solution steps

16 r^{2} + 8 r + 1 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 16 \cdot 1}{2 \cdot 16}

8^{2} - 4 \cdot 16 \cdot 1 = 0

r_{1, 2} = \frac{- 8 \pm 0}{2 \cdot 16}

r = \frac{- 8}{2 \cdot 16}

\frac{- 8}{2 \cdot 16} = - \frac{1}{4}

r = - \frac{1}{4}

The solution to the quadratic equation is:

r = - \frac{1}{4}

(x + 1) (x - 5) (x^{2} + 9) \leq 0

\frac{2 10}{3 11}

x^{2} - 6 x = 5

\frac{3}{10} + \frac{3}{10}

9 x^{2} - 18 x y + 5 y^{2}

What are the solutions to the equation 16r^2+8r+1=0 ?
The solutions to the equation 16r^2+8r+1=0 are r=-1/4
Find the zeros of 16r^2+8r+1=0
The zeros of 16r^2+8r+1=0 are r=-1/4