What are the solutions to the equation y^2-2y+(1+4pi^2)=0 ?

The solutions to the equation y^2-2y+(1+4pi^2)=0 are y=1+2ipi,y=1-2ipi

Find the zeros of y^2-2y+(1+4pi^2)=0

The zeros of y^2-2y+(1+4pi^2)=0 are y=1+2ipi,y=1-2ipi

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y^{2} - 2 y + (1 + 4 π^{2}) = 0

y = 1 + 2 iπ, y = 1 - 2 iπ

Solution steps

y^{2} - 2 y + (1 + 4 π^{2}) = 0

Refine

y^{2} - 2 y + 1 + 4 π^{2} = 0

Solve with the quadratic formula

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

Simplify

(- 2)^{2} - 4 \cdot 1 \cdot (1 + 4 π^{2}) : 4 iπ

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

Separate the solutions

y_{1} = \frac{- ( - 2 ) + 4 iπ}{2 \cdot 1}, y_{2} = \frac{- ( - 2 ) - 4 iπ}{2 \cdot 1}

y = \frac{- ( - 2 ) + 4 iπ}{2 \cdot 1} : 1 + 2 iπ

y = \frac{- ( - 2 ) - 4 iπ}{2 \cdot 1} : 1 - 2 iπ

The solutions to the quadratic equation are:

y = 1 + 2 iπ, y = 1 - 2 iπ

5 x^{2} - 10 x - 3 = 0

3 x^{2} - 18 x + 24 = - 2

3 x^{2} - 4 x - 4 = 0

x^{2} - 19 x - 90 = 0

25 π = π (x + 3)^{2}

What are the solutions to the equation y^2-2y+(1+4pi^2)=0 ?
The solutions to the equation y^2-2y+(1+4pi^2)=0 are y=1+2ipi,y=1-2ipi
Find the zeros of y^2-2y+(1+4pi^2)=0
The zeros of y^2-2y+(1+4pi^2)=0 are y=1+2ipi,y=1-2ipi