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

The solutions to the equation y^2+y+2=0 are y=-1/2+i(sqrt(7))/2 ,y=-1/2-i(sqrt(7))/2

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

The zeros of y^2+y+2=0 are y=-1/2+i(sqrt(7))/2 ,y=-1/2-i(sqrt(7))/2

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y^{2} + y + 2 = 0

y = - \frac{1}{2} + i \frac{7}{2}, y = - \frac{1}{2} - i \frac{7}{2}

Solution steps

y^{2} + y + 2 = 0

Solve with the quadratic formula

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

Simplify

1^{2} - 4 \cdot 1 \cdot 2 : 7 i

y_{1, 2} = \frac{- 1 \pm 7 i}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 1 + 7 i}{2 \cdot 1}, y_{2} = \frac{- 1 - 7 i}{2 \cdot 1}

y = \frac{- 1 + 7 i}{2 \cdot 1} : - \frac{1}{2} + i \frac{7}{2}

y = \frac{- 1 - 7 i}{2 \cdot 1} : - \frac{1}{2} - i \frac{7}{2}

The solutions to the quadratic equation are:

y = - \frac{1}{2} + i \frac{7}{2}, y = - \frac{1}{2} - i \frac{7}{2}

x^{3} + 2 x^{2} - 23 x - 60

12 s^{3} + 10 s^{2} - 8 s

\frac{4 m ^{2}}{( 5 m ^{4} ) ^{3}}

x y^{5} - x^{4} y

\frac{2}{3} (1)

What are the solutions to the equation y^2+y+2=0 ?
The solutions to the equation y^2+y+2=0 are y=-1/2+i(sqrt(7))/2 ,y=-1/2-i(sqrt(7))/2
Find the zeros of y^2+y+2=0
The zeros of y^2+y+2=0 are y=-1/2+i(sqrt(7))/2 ,y=-1/2-i(sqrt(7))/2