What are the solutions to the equation eln(e)(x^2-3)=2x ?

The solutions to the equation eln(e)(x^2-3)=2x are x=(2+sqrt(4+12e^2))/(2e),x=(2-sqrt(4+12e^2))/(2e)

Find the zeros of eln(e)(x^2-3)=2x

The zeros of eln(e)(x^2-3)=2x are x=(2+sqrt(4+12e^2))/(2e),x=(2-sqrt(4+12e^2))/(2e)

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e ln (e) (x^{2} - 3) = 2 x

x = \frac{2 + 4 + 12 e ^{2}}{2 e}, x = \frac{2 - 4 + 12 e ^{2}}{2 e}

Decimal Notation

x = 2.13856 \dots, x = - 1.40280 \dots

Solution steps

e ln (e) (x^{2} - 3) = 2 x

Expand

e ln (e) (x^{2} - 3) : e x^{2} - 3 e

e x^{2} - 3 e = 2 x

Move

2 x

to the left side

e x^{2} - 3 e - 2 x = 0

Write in the standard form

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

e x^{2} - 2 x - 3 e = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 e ( - 3 e )}{2 e}

(- 2)^{2} - 4 e (- 3 e) = 4 + 12 e^{2}

x_{1, 2} = \frac{- ( - 2 ) \pm 4 + 12 e ^{2}}{2 e}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 4 + 12 e ^{2}}{2 e}, x_{2} = \frac{- ( - 2 ) - 4 + 12 e ^{2}}{2 e}

x = \frac{- ( - 2 ) + 4 + 12 e ^{2}}{2 e} : \frac{2 + 4 + 12 e ^{2}}{2 e}

x = \frac{- ( - 2 ) - 4 + 12 e ^{2}}{2 e} : \frac{2 - 4 + 12 e ^{2}}{2 e}

The solutions to the quadratic equation are:

x = \frac{2 + 4 + 12 e ^{2}}{2 e}, x = \frac{2 - 4 + 12 e ^{2}}{2 e}

\frac{x}{6} + \frac{x ^{2}}{2} = 2 \frac{x}{3}

x, x^{2} + 2 y^{2} - y - 2 z = 0

a, (a + b) 1 + a^{2} = c^{6}

k, p^{2} + q^{2} = k^{2}

(k - 12) \cdot x^{2} + 2 (k - 12) \cdot x + 2 = 0

What are the solutions to the equation eln(e)(x^2-3)=2x ?
The solutions to the equation eln(e)(x^2-3)=2x are x=(2+sqrt(4+12e^2))/(2e),x=(2-sqrt(4+12e^2))/(2e)
Find the zeros of eln(e)(x^2-3)=2x
The zeros of eln(e)(x^2-3)=2x are x=(2+sqrt(4+12e^2))/(2e),x=(2-sqrt(4+12e^2))/(2e)