What are the solutions to the equation 25x^2+20x-4=0 ?

The solutions to the equation 25x^2+20x-4=0 are x=(2(sqrt(2)-1))/5 ,x=-(2(1+sqrt(2)))/5

Find the zeros of 25x^2+20x-4=0

The zeros of 25x^2+20x-4=0 are x=(2(sqrt(2)-1))/5 ,x=-(2(1+sqrt(2)))/5

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

x = \frac{2 ( 2 - 1 )}{5}, x = - \frac{2 ( 1 + 2 )}{5}

Decimal Notation

x = 0.16568 \dots, x = - 0.96568 \dots

Solution steps

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 25 ( - 4 )}{2 \cdot 25}

2 0^{2} - 4 \cdot 25 (- 4) = 202

x_{1, 2} = \frac{- 20 \pm 20 2}{2 \cdot 25}

Separate the solutions

x_{1} = \frac{- 20 + 20 2}{2 \cdot 25}, x_{2} = \frac{- 20 - 20 2}{2 \cdot 25}

x = \frac{- 20 + 20 2}{2 \cdot 25} : \frac{2 ( 2 - 1 )}{5}

x = \frac{- 20 - 20 2}{2 \cdot 25} : - \frac{2 ( 1 + 2 )}{5}

The solutions to the quadratic equation are:

x = \frac{2 ( 2 - 1 )}{5}, x = - \frac{2 ( 1 + 2 )}{5}

x (x + 4) = 5

- x^{2} + 6 x + 8 = 0

x^{2} - 12 x - 9 = 0

y^{2} - 2 y + 3 = 0

2 X^{2} - 5 X - 3 = 0

What are the solutions to the equation 25x^2+20x-4=0 ?
The solutions to the equation 25x^2+20x-4=0 are x=(2(sqrt(2)-1))/5 ,x=-(2(1+sqrt(2)))/5
Find the zeros of 25x^2+20x-4=0
The zeros of 25x^2+20x-4=0 are x=(2(sqrt(2)-1))/5 ,x=-(2(1+sqrt(2)))/5