What are the solutions to the equation y(y-4)=-8 ?

The solutions to the equation y(y-4)=-8 are y=2+2i,y=2-2i

Find the zeros of y(y-4)=-8

The zeros of y(y-4)=-8 are y=2+2i,y=2-2i

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y (y - 4) = - 8

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

Solution steps

y (y - 4) = - 8

Expand

y (y - 4) : y^{2} - 4 y

y^{2} - 4 y = - 8

Move

8

to the left side

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

y, 36 + y^{2} + z^{2} = 4

x, f (x - 2) = x^{2}

15 = x^{2}

(x + 1)^{2} - 8 = 0

x (21 - x) = 20

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