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

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

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

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

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

x = \frac{1}{5}, x = - \frac{1}{4}

Solution steps

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

Solve with the quadratic formula

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

1^{2} - 4 \cdot 20 (- 1) = 9

x_{1, 2} = \frac{- 1 \pm 9}{2 \cdot 20}

Separate the solutions

x_{1} = \frac{- 1 + 9}{2 \cdot 20}, x_{2} = \frac{- 1 - 9}{2 \cdot 20}

x = \frac{- 1 + 9}{2 \cdot 20} : \frac{1}{5}

x = \frac{- 1 - 9}{2 \cdot 20} : - \frac{1}{4}

The solutions to the quadratic equation are:

x = \frac{1}{5}, x = - \frac{1}{4}

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

2 x^{2} - 9 x - 5 < 0

2 x^{2} + 8 x - 64

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

2 m^{2} - 11 m - 40 = 0

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