What are the solutions to the equation y^2-4y+2y+5=0 ?

The solutions to the equation y^2-4y+2y+5=0 are y=1+2i,y=1-2i

Find the zeros of y^2-4y+2y+5=0

The zeros of y^2-4y+2y+5=0 are y=1+2i,y=1-2i

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y^{2} - 4 y + 2 y + 5 = 0

y = 1 + 2 i, y = 1 - 2 i

Solution steps

y^{2} - 4 y + 2 y + 5 = 0

Refine

y^{2} - 2 y + 5 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

Simplify

(- 2)^{2} - 4 \cdot 1 \cdot 5 : 4 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = 1 + 2 i, y = 1 - 2 i

b^{2} + 7 b + 12 = 0

x^{2} - 7 x + 12

4 x^{2} + 12 x + 10 = 0

- 2 x^{2} - 4 x + 5 = 0

t^{2} = 25

What are the solutions to the equation y^2-4y+2y+5=0 ?
The solutions to the equation y^2-4y+2y+5=0 are y=1+2i,y=1-2i
Find the zeros of y^2-4y+2y+5=0
The zeros of y^2-4y+2y+5=0 are y=1+2i,y=1-2i