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

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

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

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

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5 x^{2} - 4 x = 1

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

Decimal Notation

x = 1, x = - 0.2

Solution steps

5 x^{2} - 4 x = 1

Move

1

to the left side

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

\frac{( x + 2 ) ^{2}}{5} - \frac{( x ^{2} - 9 )}{4} = \frac{( x + 3 ) ^{2}}{2} + \frac{1}{5}

x, 4 x^{2} - 6 x y - 4 y^{2} = 15

w^{2} + w - 2 = 0

x^{2} + x - 30 = 0

3 x^{2} - 60 = 3 x

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