What are the solutions to the equation n^2-3n=-5 ?

The solutions to the equation n^2-3n=-5 are n= 3/2+i(sqrt(11))/2 ,n= 3/2-i(sqrt(11))/2

Find the zeros of n^2-3n=-5

The zeros of n^2-3n=-5 are n= 3/2+i(sqrt(11))/2 ,n= 3/2-i(sqrt(11))/2

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n^{2} - 3 n = - 5

n = \frac{3}{2} + i \frac{11}{2}, n = \frac{3}{2} - i \frac{11}{2}

Solution steps

n^{2} - 3 n = - 5

Move

5

to the left side

n^{2} - 3 n + 5 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

Simplify

(- 3)^{2} - 4 \cdot 1 \cdot 5 : 11 i

n_{1, 2} = \frac{- ( - 3 ) \pm 11 i}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 3 ) + 11 i}{2 \cdot 1}, n_{2} = \frac{- ( - 3 ) - 11 i}{2 \cdot 1}

n = \frac{- ( - 3 ) + 11 i}{2 \cdot 1} : \frac{3}{2} + i \frac{11}{2}

n = \frac{- ( - 3 ) - 11 i}{2 \cdot 1} : \frac{3}{2} - i \frac{11}{2}

The solutions to the quadratic equation are:

n = \frac{3}{2} + i \frac{11}{2}, n = \frac{3}{2} - i \frac{11}{2}

\frac{a - b}{a ^{- 1} - b ^{- 1}}

(\frac{- 3}{5})^{2}

6 x^{2} - 12 x - 18

\frac{7 π}{12} + π

9^{\frac{- 3}{2}}

What are the solutions to the equation n^2-3n=-5 ?
The solutions to the equation n^2-3n=-5 are n= 3/2+i(sqrt(11))/2 ,n= 3/2-i(sqrt(11))/2
Find the zeros of n^2-3n=-5
The zeros of n^2-3n=-5 are n= 3/2+i(sqrt(11))/2 ,n= 3/2-i(sqrt(11))/2