What are the solutions to the equation e^2+8+4e=0 ?

The solutions to the equation e^2+8+4e=0 are e=-2+2i,e=-2-2i

Find the zeros of e^2+8+4e=0

The zeros of e^2+8+4e=0 are e=-2+2i,e=-2-2i

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e^{2} + 8 + 4 e = 0

e = - 2 + 2 i, e = - 2 - 2 i

Solution steps

e^{2} + 8 + 4 e = 0

Write in the standard form

a x^{2} + b x + c = 0

e^{2} + 4 e + 8 = 0

Solve with the quadratic formula

e_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot 8}{2 \cdot 1}

Simplify

4^{2} - 4 \cdot 1 \cdot 8 : 4 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

e = - 2 + 2 i, e = - 2 - 2 i

15 x^{2} = 38 x - 24

x^{2} + 11 = 35

2 x^{2} + 5 x + 2 = 5

x (2 x - 4) + 6 = 3 x^{2} - 7 x

18 x^{2} = - 21 x

What are the solutions to the equation e^2+8+4e=0 ?
The solutions to the equation e^2+8+4e=0 are e=-2+2i,e=-2-2i
Find the zeros of e^2+8+4e=0
The zeros of e^2+8+4e=0 are e=-2+2i,e=-2-2i