What are the solutions to the equation 2/7 n^2-1/7 n+1/2 =0 ?

The solutions to the equation 2/7 n^2-1/7 n+1/2 =0 are n= 1/4+i(3sqrt(3))/4 ,n= 1/4-i(3sqrt(3))/4

Find the zeros of 2/7 n^2-1/7 n+1/2 =0

The zeros of 2/7 n^2-1/7 n+1/2 =0 are n= 1/4+i(3sqrt(3))/4 ,n= 1/4-i(3sqrt(3))/4

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\frac{2}{7} n^{2} - \frac{1}{7} n + \frac{1}{2} = 0

n = \frac{1}{4} + i \frac{3 3}{4}, n = \frac{1}{4} - i \frac{3 3}{4}

Solution steps

\frac{2}{7} n^{2} - \frac{1}{7} n + \frac{1}{2} = 0

Find Least Common Multiplier of

7, 2 : 14

Multiply by LCM=

14

\frac{2}{7} n^{2} \cdot 14 - \frac{1}{7} n \cdot 14 + \frac{1}{2} \cdot 14 = 0 \cdot 14

Simplify

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

Solve with the quadratic formula

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

Simplify

(- 2)^{2} - 4 \cdot 4 \cdot 7 : 63 i

n_{1, 2} = \frac{- ( - 2 ) \pm 6 3 i}{2 \cdot 4}

Separate the solutions

n_{1} = \frac{- ( - 2 ) + 6 3 i}{2 \cdot 4}, n_{2} = \frac{- ( - 2 ) - 6 3 i}{2 \cdot 4}

n = \frac{- ( - 2 ) + 6 3 i}{2 \cdot 4} : \frac{1}{4} + i \frac{3 3}{4}

n = \frac{- ( - 2 ) - 6 3 i}{2 \cdot 4} : \frac{1}{4} - i \frac{3 3}{4}

The solutions to the quadratic equation are:

n = \frac{1}{4} + i \frac{3 3}{4}, n = \frac{1}{4} - i \frac{3 3}{4}

\frac{15}{x + 3} > 4

\frac{3}{4} + \frac{1}{8}

\frac{f ( a + h ) - f ( a )}{h}

28 - k \geq 7 (k - 4)

9 + n \geq - 4

What are the solutions to the equation 2/7 n^2-1/7 n+1/2 =0 ?
The solutions to the equation 2/7 n^2-1/7 n+1/2 =0 are n= 1/4+i(3sqrt(3))/4 ,n= 1/4-i(3sqrt(3))/4
Find the zeros of 2/7 n^2-1/7 n+1/2 =0
The zeros of 2/7 n^2-1/7 n+1/2 =0 are n= 1/4+i(3sqrt(3))/4 ,n= 1/4-i(3sqrt(3))/4