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 = 0.17082 \dots, x = - 1.17082 \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}

\frac{x + 8}{6} + 4 > \frac{x + 7}{5} + 3

2 a^{2} + 4 a = 6

9 x^{2} + 12 x y + 4 y^{2}

2 x \cdot x

(4^{8})^{3}

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))