What are the solutions to the equation y^2-8y+32=0 ?

The solutions to the equation y^2-8y+32=0 are y=4+4i,y=4-4i

Find the zeros of y^2-8y+32=0

The zeros of y^2-8y+32=0 are y=4+4i,y=4-4i

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y^{2} - 8 y + 32 = 0

y = 4 + 4 i, y = 4 - 4 i

Solution steps

y^{2} - 8 y + 32 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot 32}{2 \cdot 1}

Simplify

(- 8)^{2} - 4 \cdot 1 \cdot 32 : 8 i

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

Separate the solutions

y_{1} = \frac{- ( - 8 ) + 8 i}{2 \cdot 1}, y_{2} = \frac{- ( - 8 ) - 8 i}{2 \cdot 1}

y = \frac{- ( - 8 ) + 8 i}{2 \cdot 1} : 4 + 4 i

y = \frac{- ( - 8 ) - 8 i}{2 \cdot 1} : 4 - 4 i

The solutions to the quadratic equation are:

y = 4 + 4 i, y = 4 - 4 i

- 3 (x + 7) = 24

e^{i \cdot k \cdot 3} + e^{- i \cdot k \cdot 3}

- 5 t^{2} + 20 t + 60

36 x + 2

x^{2} + \frac{1}{4} x - 1 = 0

What are the solutions to the equation y^2-8y+32=0 ?
The solutions to the equation y^2-8y+32=0 are y=4+4i,y=4-4i
Find the zeros of y^2-8y+32=0
The zeros of y^2-8y+32=0 are y=4+4i,y=4-4i