What are the solutions to the equation 11e^2+5=7e ?

The solutions to the equation 11e^2+5=7e are e= 7/22+i(3sqrt(19))/(22),e= 7/22-i(3sqrt(19))/(22)

Find the zeros of 11e^2+5=7e

The zeros of 11e^2+5=7e are e= 7/22+i(3sqrt(19))/(22),e= 7/22-i(3sqrt(19))/(22)

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11 e^{2} + 5 = 7 e

e = \frac{7}{22} + i \frac{3 19}{22}, e = \frac{7}{22} - i \frac{3 19}{22}

Solution steps

11 e^{2} + 5 = 7 e

Move

7 e

to the left side

11 e^{2} + 5 - 7 e = 0

Write in the standard form

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

11 e^{2} - 7 e + 5 = 0

Solve with the quadratic formula

e_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 11 \cdot 5}{2 \cdot 11}

Simplify

(- 7)^{2} - 4 \cdot 11 \cdot 5 : 319 i

e_{1, 2} = \frac{- ( - 7 ) \pm 3 19 i}{2 \cdot 11}

Separate the solutions

e_{1} = \frac{- ( - 7 ) + 3 19 i}{2 \cdot 11}, e_{2} = \frac{- ( - 7 ) - 3 19 i}{2 \cdot 11}

e = \frac{- ( - 7 ) + 3 19 i}{2 \cdot 11} : \frac{7}{22} + i \frac{3 19}{22}

e = \frac{- ( - 7 ) - 3 19 i}{2 \cdot 11} : \frac{7}{22} - i \frac{3 19}{22}

The solutions to the quadratic equation are:

e = \frac{7}{22} + i \frac{3 19}{22}, e = \frac{7}{22} - i \frac{3 19}{22}

12 x = 2 x^{2}

2 x^{2} + 13 = 15

6 x^{2} - 37 x + 6 = 0

y^{2} + 7 y - 8 = 0

\frac{1}{9} = x^{2}

What are the solutions to the equation 11e^2+5=7e ?
The solutions to the equation 11e^2+5=7e are e= 7/22+i(3sqrt(19))/(22),e= 7/22-i(3sqrt(19))/(22)
Find the zeros of 11e^2+5=7e
The zeros of 11e^2+5=7e are e= 7/22+i(3sqrt(19))/(22),e= 7/22-i(3sqrt(19))/(22)