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

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

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

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

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

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

Solution steps

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

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

1 \leq x \leq 4

7 (8 x + 5) = 3 (7 x + 3) + 166

(- 5 x^{5} y^{3})^{3}

33 = 5 x - 12

(x - 7)^{2} = 81

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