What are the solutions to the equation -19=-3-n^2-3-n^2 ?

The solutions to the equation -19=-3-n^2-3-n^2 are n=sqrt(13/2),n=-sqrt(13/2)

Find the zeros of -19=-3-n^2-3-n^2

The zeros of -19=-3-n^2-3-n^2 are n=sqrt(13/2),n=-sqrt(13/2)

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- 19 = - 3 - n^{2} - 3 - n^{2}

n = \frac{13}{2}, n = - \frac{13}{2}

Decimal Notation

n = 2.54950 \dots, n = - 2.54950 \dots

Solution steps

- 19 = - 3 - n^{2} - 3 - n^{2}

Switch sides

- 3 - n^{2} - 3 - n^{2} = - 19

Move

3

to the right side

- n^{2} - 3 - n^{2} = - 16

Move

3

to the right side

- n^{2} - n^{2} = - 13

Add similar elements:

- n^{2} - n^{2} = - 2 n^{2}

- 2 n^{2} = - 13

Divide both sides by

- 2

n^{2} = \frac{13}{2}

For

x^{2} = f (a)

the solutions are

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

n = \frac{13}{2}, n = - \frac{13}{2}

a^{2} - 3 a - 10 = 0

2 a^{2} - 14 a = 0

7 (x + 1)^{2} - 28 = 0

0 = 3 x^{2} - 8 x + 4

a = 2 a^{2} - 1

What are the solutions to the equation -19=-3-n^2-3-n^2 ?
The solutions to the equation -19=-3-n^2-3-n^2 are n=sqrt(13/2),n=-sqrt(13/2)
Find the zeros of -19=-3-n^2-3-n^2
The zeros of -19=-3-n^2-3-n^2 are n=sqrt(13/2),n=-sqrt(13/2)