What are the solutions to the equation n^2-3n=-7 ?

The solutions to the equation n^2-3n=-7 are n= 3/2+i(sqrt(19))/2 ,n= 3/2-i(sqrt(19))/2

Find the zeros of n^2-3n=-7

The zeros of n^2-3n=-7 are n= 3/2+i(sqrt(19))/2 ,n= 3/2-i(sqrt(19))/2

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n^{2} - 3 n = - 7

n = \frac{3}{2} + i \frac{19}{2}, n = \frac{3}{2} - i \frac{19}{2}

Solution steps

n^{2} - 3 n = - 7

Move

7

to the left side

n^{2} - 3 n + 7 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot 7}{2 \cdot 1}

Simplify

(- 3)^{2} - 4 \cdot 1 \cdot 7 : 19 i

n_{1, 2} = \frac{- ( - 3 ) \pm 19 i}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 3 ) + 19 i}{2 \cdot 1}, n_{2} = \frac{- ( - 3 ) - 19 i}{2 \cdot 1}

n = \frac{- ( - 3 ) + 19 i}{2 \cdot 1} : \frac{3}{2} + i \frac{19}{2}

n = \frac{- ( - 3 ) - 19 i}{2 \cdot 1} : \frac{3}{2} - i \frac{19}{2}

The solutions to the quadratic equation are:

n = \frac{3}{2} + i \frac{19}{2}, n = \frac{3}{2} - i \frac{19}{2}

- 2 x^{2} + x + 1

- x = 1

\frac{x}{9} = 81

\frac{1}{4} (x - 4) + \frac{2}{5} (x - 5) = \frac{1}{5} (x^{2} - 53)

x - \frac{2}{3} > 2 x + \frac{4}{3}

What are the solutions to the equation n^2-3n=-7 ?
The solutions to the equation n^2-3n=-7 are n= 3/2+i(sqrt(19))/2 ,n= 3/2-i(sqrt(19))/2
Find the zeros of n^2-3n=-7
The zeros of n^2-3n=-7 are n= 3/2+i(sqrt(19))/2 ,n= 3/2-i(sqrt(19))/2