What are the solutions to the equation (n+1)^2=sqrt(5)+2n ?

The solutions to the equation (n+1)^2=sqrt(5)+2n are n=sqrt(\sqrt{5)-1},n=-sqrt(\sqrt{5)-1}

Find the zeros of (n+1)^2=sqrt(5)+2n

The zeros of (n+1)^2=sqrt(5)+2n are n=sqrt(\sqrt{5)-1},n=-sqrt(\sqrt{5)-1}

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(n + 1)^{2} = 5 + 2 n

n = 5 - 1, n = - 5 - 1

Decimal Notation

n = 1.11178 \dots, n = - 1.11178 \dots

Solution steps

(n + 1)^{2} = 5 + 2 n

Expand

(n + 1)^{2} : n^{2} + 2 n + 1

n^{2} + 2 n + 1 = 5 + 2 n

Move

1

to the right side

n^{2} + 2 n = 5 + 2 n - 1

Move

2 n

to the left side

n^{2} = 5 - 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

n = 5 - 1, n = - 5 - 1

0 = 4.8 y^{2} - 1428 y + 44100

- 9 = x^{2}

t^{2} = - 63 t + 650

3 r^{2} + 8 r + 6 = 0

x, y^{1} + 3 x^{2} y = x^{2}

What are the solutions to the equation (n+1)^2=sqrt(5)+2n ?
The solutions to the equation (n+1)^2=sqrt(5)+2n are n=sqrt(\sqrt{5)-1},n=-sqrt(\sqrt{5)-1}
Find the zeros of (n+1)^2=sqrt(5)+2n
The zeros of (n+1)^2=sqrt(5)+2n are n=sqrt(\sqrt{5)-1},n=-sqrt(\sqrt{5)-1}