What are the solutions to the equation -4x^2-3x+1=0 ?

The solutions to the equation -4x^2-3x+1=0 are x=-1,x= 1/4

Find the zeros of -4x^2-3x+1=0

The zeros of -4x^2-3x+1=0 are x=-1,x= 1/4

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- 4 x^{2} - 3 x + 1 = 0

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

Solution steps

- 4 x^{2} - 3 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 4 ) \cdot 1}{2 ( - 4 )}

(- 3)^{2} - 4 (- 4) \cdot 1 = 5

x_{1, 2} = \frac{- ( - 3 ) \pm 5}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 5}{2 ( - 4 )}, x_{2} = \frac{- ( - 3 ) - 5}{2 ( - 4 )}

x = \frac{- ( - 3 ) + 5}{2 ( - 4 )} : - 1

x = \frac{- ( - 3 ) - 5}{2 ( - 4 )} : \frac{1}{4}

The solutions to the quadratic equation are:

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

(2 + 5) (3 + 5)

\frac{9 - 2 x}{5} = 9

x^{4} - 6 x^{2} - 27

(2 + 4 i)^{2}

3 \cdot \frac{1}{3}

What are the solutions to the equation -4x^2-3x+1=0 ?
The solutions to the equation -4x^2-3x+1=0 are x=-1,x= 1/4
Find the zeros of -4x^2-3x+1=0
The zeros of -4x^2-3x+1=0 are x=-1,x= 1/4