What are the solutions to the equation x^2+2x+n^2=0 ?

The solutions to the equation x^2+2x+n^2=0 are x=-1+sqrt(-n^2+1),x=-sqrt(-n^2+1)-1

Find the zeros of x^2+2x+n^2=0

The zeros of x^2+2x+n^2=0 are x=-1+sqrt(-n^2+1),x=-sqrt(-n^2+1)-1

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x^{2} + 2 x + n^{2} = 0

x = - 1 + - n^{2} + 1, x = - - n^{2} + 1 - 1

Solution steps

x^{2} + 2 x + n^{2} = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot n ^{2}}{2 \cdot 1}

Simplify

2^{2} - 4 \cdot 1 \cdot n^{2} : 2 1 - n^{2}

x_{1, 2} = \frac{- 2 \pm 2 1 - n ^{2}}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 2 1 - n ^{2}}{2 \cdot 1}, x_{2} = \frac{- 2 - 2 1 - n ^{2}}{2 \cdot 1}

x = \frac{- 2 + 2 1 - n ^{2}}{2 \cdot 1} : - 1 + - n^{2} + 1

x = \frac{- 2 - 2 1 - n ^{2}}{2 \cdot 1} : - - n^{2} + 1 - 1

The solutions to the quadratic equation are:

x = - 1 + - n^{2} + 1, x = - - n^{2} + 1 - 1

- 16 t^{2} + 32 t + 384 = 0

y (y + 5) = 750

4 x^{2} + 16 x + 13 = 0

x^{2} - 20 x + 33 = 0

4 x^{2} - 20 = 5

What are the solutions to the equation x^2+2x+n^2=0 ?
The solutions to the equation x^2+2x+n^2=0 are x=-1+sqrt(-n^2+1),x=-sqrt(-n^2+1)-1
Find the zeros of x^2+2x+n^2=0
The zeros of x^2+2x+n^2=0 are x=-1+sqrt(-n^2+1),x=-sqrt(-n^2+1)-1