What are the solutions to the equation 4n^2-2n=-4 ?

The solutions to the equation 4n^2-2n=-4 are n= 1/4+i(sqrt(15))/4 ,n= 1/4-i(sqrt(15))/4

Find the zeros of 4n^2-2n=-4

The zeros of 4n^2-2n=-4 are n= 1/4+i(sqrt(15))/4 ,n= 1/4-i(sqrt(15))/4

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4 n^{2} - 2 n = - 4

n = \frac{1}{4} + i \frac{15}{4}, n = \frac{1}{4} - i \frac{15}{4}

Solution steps

4 n^{2} - 2 n = - 4

Move

4

to the left side

4 n^{2} - 2 n + 4 = 0

Solve with the quadratic formula

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

Simplify

(- 2)^{2} - 4 \cdot 4 \cdot 4 : 215 i

n_{1, 2} = \frac{- ( - 2 ) \pm 2 15 i}{2 \cdot 4}

Separate the solutions

n_{1} = \frac{- ( - 2 ) + 2 15 i}{2 \cdot 4}, n_{2} = \frac{- ( - 2 ) - 2 15 i}{2 \cdot 4}

n = \frac{- ( - 2 ) + 2 15 i}{2 \cdot 4} : \frac{1}{4} + i \frac{15}{4}

n = \frac{- ( - 2 ) - 2 15 i}{2 \cdot 4} : \frac{1}{4} - i \frac{15}{4}

The solutions to the quadratic equation are:

n = \frac{1}{4} + i \frac{15}{4}, n = \frac{1}{4} - i \frac{15}{4}

v, a = \frac{v ^{2}}{r}

(19 x - 3) (9 + x) = 0

5 x^{2} + 33 = 3

- x^{2} - 8 x + 12 = 0

81 x^{2} + 36 x + 4 = 0

What are the solutions to the equation 4n^2-2n=-4 ?
The solutions to the equation 4n^2-2n=-4 are n= 1/4+i(sqrt(15))/4 ,n= 1/4-i(sqrt(15))/4
Find the zeros of 4n^2-2n=-4
The zeros of 4n^2-2n=-4 are n= 1/4+i(sqrt(15))/4 ,n= 1/4-i(sqrt(15))/4