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

The solutions to the equation 5x^2-5x=1 are x=(sqrt(5)+3)/(2sqrt(5)),x=(sqrt(5)-3)/(2sqrt(5))

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

The zeros of 5x^2-5x=1 are x=(sqrt(5)+3)/(2sqrt(5)),x=(sqrt(5)-3)/(2sqrt(5))

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5 x^{2} - 5 x = 1

x = \frac{5 + 3}{2 5}, x = \frac{5 - 3}{2 5}

Decimal Notation

x = 1.17082 \dots, x = - 0.17082 \dots

Solution steps

5 x^{2} - 5 x = 1

Move

1

to the left side

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

Solve with the quadratic formula

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

(- 5)^{2} - 4 \cdot 5 (- 1) = 35

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = \frac{5 + 3}{2 5}, x = \frac{5 - 3}{2 5}

(3 + i)^{2}

\frac{x}{4} = - 2

- 7.687 - 1.74

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

\frac{2}{2 + k} < 1

What are the solutions to the equation 5x^2-5x=1 ?
The solutions to the equation 5x^2-5x=1 are x=(sqrt(5)+3)/(2sqrt(5)),x=(sqrt(5)-3)/(2sqrt(5))
Find the zeros of 5x^2-5x=1
The zeros of 5x^2-5x=1 are x=(sqrt(5)+3)/(2sqrt(5)),x=(sqrt(5)-3)/(2sqrt(5))