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

The solutions to the equation y^2-4y+12+4=0 are y=2+2sqrt(3)i,y=2-2sqrt(3)i

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

The zeros of y^2-4y+12+4=0 are y=2+2sqrt(3)i,y=2-2sqrt(3)i

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

y = 2 + 23 i, y = 2 - 23 i

Solution steps

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

Refine

y^{2} - 4 y + 16 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 16}{2 \cdot 1}

Simplify

(- 4)^{2} - 4 \cdot 1 \cdot 16 : 43 i

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

Separate the solutions

y_{1} = \frac{- ( - 4 ) + 4 3 i}{2 \cdot 1}, y_{2} = \frac{- ( - 4 ) - 4 3 i}{2 \cdot 1}

y = \frac{- ( - 4 ) + 4 3 i}{2 \cdot 1} : 2 + 23 i

y = \frac{- ( - 4 ) - 4 3 i}{2 \cdot 1} : 2 - 23 i

The solutions to the quadratic equation are:

y = 2 + 23 i, y = 2 - 23 i

2 x^{2} - 3 x - 1 = (x + 3) (x + 4)

- 16 x^{2} = 0

x^{2} + 3 x - 48 = 0

4 x^{2} = 1, 2 x + 9 = 0

t^{2} + 14 t + 45 = 0

What are the solutions to the equation y^2-4y+12+4=0 ?
The solutions to the equation y^2-4y+12+4=0 are y=2+2sqrt(3)i,y=2-2sqrt(3)i
Find the zeros of y^2-4y+12+4=0
The zeros of y^2-4y+12+4=0 are y=2+2sqrt(3)i,y=2-2sqrt(3)i