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

The solutions to the equation y^2+y+5=0 are y=-1/2+i(sqrt(19))/2 ,y=-1/2-i(sqrt(19))/2

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

The zeros of y^2+y+5=0 are y=-1/2+i(sqrt(19))/2 ,y=-1/2-i(sqrt(19))/2

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y^{2} + y + 5 = 0

y = - \frac{1}{2} + i \frac{19}{2}, y = - \frac{1}{2} - i \frac{19}{2}

Solution steps

y^{2} + y + 5 = 0

Solve with the quadratic formula

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

Simplify

1^{2} - 4 \cdot 1 \cdot 5 : 19 i

y_{1, 2} = \frac{- 1 \pm 19 i}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 1 + 19 i}{2 \cdot 1}, y_{2} = \frac{- 1 - 19 i}{2 \cdot 1}

y = \frac{- 1 + 19 i}{2 \cdot 1} : - \frac{1}{2} + i \frac{19}{2}

y = \frac{- 1 - 19 i}{2 \cdot 1} : - \frac{1}{2} - i \frac{19}{2}

The solutions to the quadratic equation are:

y = - \frac{1}{2} + i \frac{19}{2}, y = - \frac{1}{2} - i \frac{19}{2}

4 (x - 3) - 8 \leq 5 - x

x^{3} + 2 x^{2} - 25 x - 50

\frac{x - 7}{- x - 2} < 0

∣ 2 x + 3 ∣ \leq - 2

7^{2} + 6^{2}

What are the solutions to the equation y^2+y+5=0 ?
The solutions to the equation y^2+y+5=0 are y=-1/2+i(sqrt(19))/2 ,y=-1/2-i(sqrt(19))/2
Find the zeros of y^2+y+5=0
The zeros of y^2+y+5=0 are y=-1/2+i(sqrt(19))/2 ,y=-1/2-i(sqrt(19))/2