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

The solutions to the equation 4v^2+4v+1=0 are v=-1/2

Find the zeros of 4v^2+4v+1=0

The zeros of 4v^2+4v+1=0 are v=-1/2

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4 v^{2} + 4 v + 1 = 0

v = - \frac{1}{2}

Decimal Notation

v = - 0.5

Solution steps

4 v^{2} + 4 v + 1 = 0

Solve with the quadratic formula

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

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

v_{1, 2} = \frac{- 4 \pm 0}{2 \cdot 4}

v = \frac{- 4}{2 \cdot 4}

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

v = - \frac{1}{2}

The solution to the quadratic equation is:

v = - \frac{1}{2}

(x - 5) (x + 1) + 5 = 0

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

7 = - x^{2} - 9 x

x^{2} - \frac{3}{4} x - \frac{27}{64} = 0

x, 9 x^{2} + y^{2} + 4 z^{2} - 18 x + 2 y + 16 z = 10

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