What are the solutions to the equation y^2-4y+12=0 ?

The solutions to the equation y^2-4y+12=0 are y=2+2sqrt(2)i,y=2-2sqrt(2)i

Find the zeros of y^2-4y+12=0

The zeros of y^2-4y+12=0 are y=2+2sqrt(2)i,y=2-2sqrt(2)i

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y^{2} - 4 y + 12 = 0

y = 2 + 22 i, y = 2 - 22 i

Solution steps

y^{2} - 4 y + 12 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

Simplify

(- 4)^{2} - 4 \cdot 1 \cdot 12 : 42 i

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

Separate the solutions

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

y = \frac{- ( - 4 ) + 4 2 i}{2 \cdot 1} : 2 + 22 i

y = \frac{- ( - 4 ) - 4 2 i}{2 \cdot 1} : 2 - 22 i

The solutions to the quadratic equation are:

y = 2 + 22 i, y = 2 - 22 i

x^{2} = 6^{2} + 6^{2}

x^{2} - 200 x + 9900 = 0

s^{2} + s - 1 = 0

- 1 = x^{2} + 2 x

- x^{2} + 18 x - 80 = 0

What are the solutions to the equation y^2-4y+12=0 ?
The solutions to the equation y^2-4y+12=0 are y=2+2sqrt(2)i,y=2-2sqrt(2)i
Find the zeros of y^2-4y+12=0
The zeros of y^2-4y+12=0 are y=2+2sqrt(2)i,y=2-2sqrt(2)i