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

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

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

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

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2 x^{2} + x - \frac{1}{4} = 0

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

Solution steps

2 x^{2} + x - \frac{1}{4} = 0

Multiply both sides by

4

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

2 (\frac{1}{2})^{1}

w^{2} + 3 = 19

y^{2} + 3 y + \frac{9}{4}

m^{2} + m + 2 = 0

(2 x + 4) \cdot (x - 1) + (3 x + 5)^{2} = 3 (2 x + 5)^{2} + x

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