What are the solutions to the equation y^2+3=2y ?

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

Find the zeros of y^2+3=2y

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

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y^{2} + 3 = 2 y

y = 1 + 2 i, y = 1 - 2 i

Solution steps

y^{2} + 3 = 2 y

Move

2 y

to the left side

y^{2} + 3 - 2 y = 0

Write in the standard form

a x^{2} + b x + c = 0

y^{2} - 2 y + 3 = 0

Solve with the quadratic formula

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

Simplify

(- 2)^{2} - 4 \cdot 1 \cdot 3 : 22 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = 1 + 2 i, y = 1 - 2 i

\frac{18 x - 6 x}{6 x y}

(6^{2})^{4}

5 \leq - 5 x + 5 < 15

x (x - 9) = 0

7 ∣ 2 x - 3 ∣ = 21

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