What are the solutions to the equation 5x^2-20x+4=0 ?

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

Find the zeros of 5x^2-20x+4=0

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

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

x = \frac{2 ( 5 + 2 5 )}{5}, x = \frac{2 ( 5 - 2 5 )}{5}

Decimal Notation

x = 3.78885 \dots, x = 0.21114 \dots

Solution steps

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 5 \cdot 4}{2 \cdot 5}

(- 20)^{2} - 4 \cdot 5 \cdot 4 = 85

x_{1, 2} = \frac{- ( - 20 ) \pm 8 5}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 20 ) + 8 5}{2 \cdot 5}, x_{2} = \frac{- ( - 20 ) - 8 5}{2 \cdot 5}

x = \frac{- ( - 20 ) + 8 5}{2 \cdot 5} : \frac{2 ( 5 + 2 5 )}{5}

x = \frac{- ( - 20 ) - 8 5}{2 \cdot 5} : \frac{2 ( 5 - 2 5 )}{5}

The solutions to the quadratic equation are:

x = \frac{2 ( 5 + 2 5 )}{5}, x = \frac{2 ( 5 - 2 5 )}{5}

4 x^{2} - 10 = 278

(j - 2) j - \frac{j + 2}{3} - \frac{( j - 2 ) ( j + 2 )}{2} = (j - 2)^{2} - 4

n^{2} + n = 30

r^{2} + 6 = 7

x^{2} + 15 x - 56 = 0

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