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

The solutions to the equation n^2-3=-2n are n=1,n=-3

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

The zeros of n^2-3=-2n are n=1,n=-3

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n^{2} - 3 = - 2 n

n = 1, n = - 3

Solution steps

n^{2} - 3 = - 2 n

Move

2 n

to the left side

n^{2} - 3 + 2 n = 0

Write in the standard form

a x^{2} + b x + c = 0

n^{2} + 2 n - 3 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 3) = 4

n_{1, 2} = \frac{- 2 \pm 4}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- 2 + 4}{2 \cdot 1}, n_{2} = \frac{- 2 - 4}{2 \cdot 1}

n = \frac{- 2 + 4}{2 \cdot 1} : 1

n = \frac{- 2 - 4}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

n = 1, n = - 3

\frac{u ^{\frac{5}{8}}}{u ^{\frac{3}{4}}}

4 < n - 9

a^{3} - 2 a^{2} - a + 2

\frac{8}{3} + 1

x^{3} < 4 x

What are the solutions to the equation n^2-3=-2n ?
The solutions to the equation n^2-3=-2n are n=1,n=-3
Find the zeros of n^2-3=-2n
The zeros of n^2-3=-2n are n=1,n=-3