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 Notation

x = 4.94948 \dots, x = 0.05051 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 20)^{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}

t^{2} - 8 t - 5 = 0

x^{2} + 6.5 x + 10 = 0

x^{2} + 4 x + 18 = 0

x^{2} - 10 x + 25 = 49

20 = - 2 x^{2} + 13 x

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