What are the solutions to the equation -5x^2-4x+1=0 ?

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

Find the zeros of -5x^2-4x+1=0

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

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

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

Solution steps

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

Solve with the quadratic formula

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

(- 4)^{2} - 4 (- 5) \cdot 1 = 6

x_{1, 2} = \frac{- ( - 4 ) \pm 6}{2 ( - 5 )}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 6}{2 ( - 5 )}, x_{2} = \frac{- ( - 4 ) - 6}{2 ( - 5 )}

x = \frac{- ( - 4 ) + 6}{2 ( - 5 )} : - 1

x = \frac{- ( - 4 ) - 6}{2 ( - 5 )} : \frac{1}{5}

The solutions to the quadratic equation are:

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

6 x + 1 = 0

6 d (d + 8) = 0

\frac{3 t}{2 - x} + \frac{5}{x - 2}

- 3 (2 x^{4}) (4 x^{5} y)^{2}

4 x + 8 = 3 x + 17

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