What are the solutions to the equation 11n^2-11n=-3 ?

The solutions to the equation 11n^2-11n=-3 are n= 1/2+i(sqrt(11))/(22),n= 1/2-i(sqrt(11))/(22)

Find the zeros of 11n^2-11n=-3

The zeros of 11n^2-11n=-3 are n= 1/2+i(sqrt(11))/(22),n= 1/2-i(sqrt(11))/(22)

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11 n^{2} - 11 n = - 3

n = \frac{1}{2} + i \frac{11}{22}, n = \frac{1}{2} - i \frac{11}{22}

Solution steps

11 n^{2} - 11 n = - 3

Move

3

to the left side

11 n^{2} - 11 n + 3 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 11 \cdot 3}{2 \cdot 11}

Simplify

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

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

Separate the solutions

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

n = \frac{- ( - 11 ) + 11 i}{2 \cdot 11} : \frac{1}{2} + i \frac{11}{22}

n = \frac{- ( - 11 ) - 11 i}{2 \cdot 11} : \frac{1}{2} - i \frac{11}{22}

The solutions to the quadratic equation are:

n = \frac{1}{2} + i \frac{11}{22}, n = \frac{1}{2} - i \frac{11}{22}

2 w^{2} - 7 w - 4 = 0

a, c^{2} = a^{2} + a^{2}

2 y^{2} - 4 y + 1 = 0

8.1 \cdot 1 0^{- 12} = (2 x)^{2} \cdot (2.5 \cdot 1 0^{- 3})

Q, p = 6 - 6 Q + 3 Q^{2}

What are the solutions to the equation 11n^2-11n=-3 ?
The solutions to the equation 11n^2-11n=-3 are n= 1/2+i(sqrt(11))/(22),n= 1/2-i(sqrt(11))/(22)
Find the zeros of 11n^2-11n=-3
The zeros of 11n^2-11n=-3 are n= 1/2+i(sqrt(11))/(22),n= 1/2-i(sqrt(11))/(22)