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

The solutions to the equation 4x^2+2x-4=0 are x=(-1+sqrt(17))/4 ,x=-(1+sqrt(17))/4

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

The zeros of 4x^2+2x-4=0 are x=(-1+sqrt(17))/4 ,x=-(1+sqrt(17))/4

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

x = \frac{- 1 + 17}{4}, x = - \frac{1 + 17}{4}

Solution steps

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 4 (- 4) = 217

x_{1, 2} = \frac{- 2 \pm 2 17}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 2 + 2 17}{2 \cdot 4}, x_{2} = \frac{- 2 - 2 17}{2 \cdot 4}

x = \frac{- 2 + 2 17}{2 \cdot 4} : \frac{- 1 + 17}{4}

x = \frac{- 2 - 2 17}{2 \cdot 4} : - \frac{1 + 17}{4}

The solutions to the quadratic equation are:

x = \frac{- 1 + 17}{4}, x = - \frac{1 + 17}{4}

2 x^{2} = - 4 x + 6

11025

\frac{x}{15} \leq 450

\frac{3 n}{5} + \frac{1}{2} = \frac{5 n - 1}{10}

∣ - 2 ∣ x = 4

What are the solutions to the equation 4x^2+2x-4=0 ?
The solutions to the equation 4x^2+2x-4=0 are x=(-1+sqrt(17))/4 ,x=-(1+sqrt(17))/4
Find the zeros of 4x^2+2x-4=0
The zeros of 4x^2+2x-4=0 are x=(-1+sqrt(17))/4 ,x=-(1+sqrt(17))/4