What are the solutions to the equation y^2-4y+8=0 ?

The solutions to the equation y^2-4y+8=0 are y=2+2i,y=2-2i

Find the zeros of y^2-4y+8=0

The zeros of y^2-4y+8=0 are y=2+2i,y=2-2i

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y^{2} - 4 y + 8 = 0

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

Solution steps

y^{2} - 4 y + 8 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 8}{2 \cdot 1}

Simplify

(- 4)^{2} - 4 \cdot 1 \cdot 8 : 4 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

5 x > - 20

x + h \cdot x

10

(x + 2)^{2} - 3 = 1

(1 + \frac{2}{n})^{n}

What are the solutions to the equation y^2-4y+8=0 ?
The solutions to the equation y^2-4y+8=0 are y=2+2i,y=2-2i
Find the zeros of y^2-4y+8=0
The zeros of y^2-4y+8=0 are y=2+2i,y=2-2i