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

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

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

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

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

y = 1 + 3 i, y = 1 - 3 i

Solution steps

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

Solve with the quadratic formula

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

Simplify

(- 2)^{2} - 4 \cdot 1 \cdot 4 : 23 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = 1 + 3 i, y = 1 - 3 i

∣ lo g_{5} (x) ∣ = 2

(- 32 y^{15})^{\frac{1}{5}}

2 x^{2} - 4 x - 6 = 90

c^{4} + 9 c^{2} + 20

616

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