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

The solutions to the equation 2e^2+8=e are e= 1/4+i(3sqrt(7))/4 ,e= 1/4-i(3sqrt(7))/4

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

The zeros of 2e^2+8=e are e= 1/4+i(3sqrt(7))/4 ,e= 1/4-i(3sqrt(7))/4

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2 e^{2} + 8 = e

e = \frac{1}{4} + i \frac{3 7}{4}, e = \frac{1}{4} - i \frac{3 7}{4}

Solution steps

2 e^{2} + 8 = e

Move

e

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 \cdot 2 \cdot 8 : 37 i

e_{1, 2} = \frac{- ( - 1 ) \pm 3 7 i}{2 \cdot 2}

Separate the solutions

e_{1} = \frac{- ( - 1 ) + 3 7 i}{2 \cdot 2}, e_{2} = \frac{- ( - 1 ) - 3 7 i}{2 \cdot 2}

e = \frac{- ( - 1 ) + 3 7 i}{2 \cdot 2} : \frac{1}{4} + i \frac{3 7}{4}

e = \frac{- ( - 1 ) - 3 7 i}{2 \cdot 2} : \frac{1}{4} - i \frac{3 7}{4}

The solutions to the quadratic equation are:

e = \frac{1}{4} + i \frac{3 7}{4}, e = \frac{1}{4} - i \frac{3 7}{4}

11 x^{2} + 1 = 2 x^{2}

2 y^{2} - 6 y + 3 = 0

6 x^{2} - 29 x + 20 = 0

(m^{2} + 4) = 0

- \frac{1}{5} x^{2} + 4 = - 1

What are the solutions to the equation 2e^2+8=e ?
The solutions to the equation 2e^2+8=e are e= 1/4+i(3sqrt(7))/4 ,e= 1/4-i(3sqrt(7))/4
Find the zeros of 2e^2+8=e
The zeros of 2e^2+8=e are e= 1/4+i(3sqrt(7))/4 ,e= 1/4-i(3sqrt(7))/4