What are the solutions to the equation x^2+((2x)/3)^2-4=0 ?

The solutions to the equation x^2+((2x)/3)^2-4=0 are x= 6/(sqrt(13)),x=-6/(sqrt(13))

Find the zeros of x^2+((2x)/3)^2-4=0

The zeros of x^2+((2x)/3)^2-4=0 are x= 6/(sqrt(13)),x=-6/(sqrt(13))

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x^{2} + (\frac{2 x}{3})^{2} - 4 = 0

x = \frac{6}{13}, x = - \frac{6}{13}

Decimal Notation

x = 1.66410 \dots, x = - 1.66410 \dots

Solution steps

x^{2} + (\frac{2 x}{3})^{2} - 4 = 0

Expand

x^{2} + (\frac{2 x}{3})^{2} - 4 : \frac{13 x ^{2}}{9} - 4

\frac{13 x ^{2}}{9} - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot \frac{13}{9} ( - 4 )}{2 \cdot \frac{13}{9}}

0^{2} - 4 \cdot \frac{13}{9} (- 4) = \frac{4 13}{3}

x_{1, 2} = \frac{- 0 \pm \frac{4 13}{3}}{2 \cdot \frac{13}{9}}

Separate the solutions

x_{1} = \frac{- 0 + \frac{4 13}{3}}{2 \cdot \frac{13}{9}}, x_{2} = \frac{- 0 - \frac{4 13}{3}}{2 \cdot \frac{13}{9}}

x = \frac{- 0 + \frac{4 13}{3}}{2 \frac{13}{9}} : \frac{6}{13}

x = \frac{- 0 - \frac{4 13}{3}}{2 \frac{13}{9}} : - \frac{6}{13}

The solutions to the quadratic equation are:

x = \frac{6}{13}, x = - \frac{6}{13}

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

3 x^{2} + 7 x

x, x^{2} + 4 x y - 3 y^{2} + x + 2 y - 6 = 0

\frac{k ^{2}}{1.2} - 7 = 23

8 x - 65 = x^{2}

What are the solutions to the equation x^2+((2x)/3)^2-4=0 ?
The solutions to the equation x^2+((2x)/3)^2-4=0 are x= 6/(sqrt(13)),x=-6/(sqrt(13))
Find the zeros of x^2+((2x)/3)^2-4=0
The zeros of x^2+((2x)/3)^2-4=0 are x= 6/(sqrt(13)),x=-6/(sqrt(13))