What are the solutions to the equation 4x^2+20x+1=0 ?

The solutions to the equation 4x^2+20x+1=0 are x=(-5+2sqrt(6))/2 ,x=-(5+2sqrt(6))/2

Find the zeros of 4x^2+20x+1=0

The zeros of 4x^2+20x+1=0 are x=(-5+2sqrt(6))/2 ,x=-(5+2sqrt(6))/2

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4 x^{2} + 20 x + 1 = 0

x = \frac{- 5 + 2 6}{2}, x = - \frac{5 + 2 6}{2}

Decimal

x = - 0.05051 \dots, x = - 4.94948 \dots

Solution steps

4 x^{2} + 20 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 4 \cdot 1}{2 \cdot 4}

2 0^{2} - 4 \cdot 4 \cdot 1 = 86

x_{1, 2} = \frac{- 20 \pm 8 6}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 20 + 8 6}{2 \cdot 4}, x_{2} = \frac{- 20 - 8 6}{2 \cdot 4}

x = \frac{- 20 + 8 6}{2 \cdot 4} : \frac{- 5 + 2 6}{2}

x = \frac{- 20 - 8 6}{2 \cdot 4} : - \frac{5 + 2 6}{2}

The solutions to the quadratic equation are:

x = \frac{- 5 + 2 6}{2}, x = - \frac{5 + 2 6}{2}

400 = - 16 t^{2} + 800

3 \frac{1}{5} x^{2} \frac{5}{8} = 8 \frac{2}{5}

3 x^{2} + 10 x = - 8

x^{2} = 4 x - 8

x^{2} + 3 x = - 7

What are the solutions to the equation 4x^2+20x+1=0 ?
The solutions to the equation 4x^2+20x+1=0 are x=(-5+2sqrt(6))/2 ,x=-(5+2sqrt(6))/2
Find the zeros of 4x^2+20x+1=0
The zeros of 4x^2+20x+1=0 are x=(-5+2sqrt(6))/2 ,x=-(5+2sqrt(6))/2