What are the solutions to the equation 4y^2-y+4=0 ?

The solutions to the equation 4y^2-y+4=0 are y= 1/8+i(3sqrt(7))/8 ,y= 1/8-i(3sqrt(7))/8

Find the zeros of 4y^2-y+4=0

The zeros of 4y^2-y+4=0 are y= 1/8+i(3sqrt(7))/8 ,y= 1/8-i(3sqrt(7))/8

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

y = \frac{1}{8} + i \frac{3 7}{8}, y = \frac{1}{8} - i \frac{3 7}{8}

Solution steps

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

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 \cdot 4 \cdot 4 : 37 i

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

Separate the solutions

y_{1} = \frac{- ( - 1 ) + 3 7 i}{2 \cdot 4}, y_{2} = \frac{- ( - 1 ) - 3 7 i}{2 \cdot 4}

y = \frac{- ( - 1 ) + 3 7 i}{2 \cdot 4} : \frac{1}{8} + i \frac{3 7}{8}

y = \frac{- ( - 1 ) - 3 7 i}{2 \cdot 4} : \frac{1}{8} - i \frac{3 7}{8}

The solutions to the quadratic equation are:

y = \frac{1}{8} + i \frac{3 7}{8}, y = \frac{1}{8} - i \frac{3 7}{8}

9 m^{2} = - 3

(x + 4)^{2} - 16 (x + 4) + 63 = 0

c^{2} = (6)^{2} + x^{2}

x, f (x + 2) = 3 x^{2} + 4

x, f (x + 2) = 3 x^{2} + 5

What are the solutions to the equation 4y^2-y+4=0 ?
The solutions to the equation 4y^2-y+4=0 are y= 1/8+i(3sqrt(7))/8 ,y= 1/8-i(3sqrt(7))/8
Find the zeros of 4y^2-y+4=0
The zeros of 4y^2-y+4=0 are y= 1/8+i(3sqrt(7))/8 ,y= 1/8-i(3sqrt(7))/8