What are the solutions to the equation x^2+((5-2x)/3)^2=2 ?

The solutions to the equation x^2+((5-2x)/3)^2=2 are x=1,x= 7/13

Find the zeros of x^2+((5-2x)/3)^2=2

The zeros of x^2+((5-2x)/3)^2=2 are x=1,x= 7/13

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x^{2} + (\frac{5 - 2 x}{3})^{2} = 2

x = 1, x = \frac{7}{13}

Decimal Notation

x = 1, x = 0.53846 \dots

Solution steps

x^{2} + (\frac{5 - 2 x}{3})^{2} = 2

Expand

x^{2} + (\frac{5 - 2 x}{3})^{2} : \frac{25}{9} - \frac{20 x}{9} + \frac{13 x ^{2}}{9}

\frac{25}{9} - \frac{20 x}{9} + \frac{13 x ^{2}}{9} = 2

Move

2

to the left side

\frac{25 - 20 x + 13 x ^{2}}{9} - 2 = 0

Write in the standard form

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

\frac{13 x ^{2}}{9} - \frac{20 x}{9} + \frac{25}{9} - 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - \frac{20}{9} ) \pm ( - \frac{20}{9} ) ^{2} - 4 \cdot \frac{13}{9} ( \frac{25}{9} - 2 )}{2 \cdot \frac{13}{9}}

(- \frac{20}{9})^{2} - 4 \cdot \frac{13}{9} (\frac{25}{9} - 2) = \frac{2}{3}

x_{1, 2} = \frac{- ( - \frac{20}{9} ) \pm \frac{2}{3}}{2 \cdot \frac{13}{9}}

Separate the solutions

x_{1} = \frac{- ( - \frac{20}{9} ) + \frac{2}{3}}{2 \cdot \frac{13}{9}}, x_{2} = \frac{- ( - \frac{20}{9} ) - \frac{2}{3}}{2 \cdot \frac{13}{9}}

x = \frac{- ( - \frac{20}{9} ) + \frac{2}{3}}{2 \frac{13}{9}} : 1

x = \frac{- ( - \frac{20}{9} ) - \frac{2}{3}}{2 \frac{13}{9}} : \frac{7}{13}

The solutions to the quadratic equation are:

x = 1, x = \frac{7}{13}

3 x^{2} - 24 x + 12 = 0

- x^{2} + 4 x + 96 = 0

9 x^{2} + 7 x + 3 = 0

x^{2} + 14 x + 49 = 16

4 x - 5 x^{2} = - 12

What are the solutions to the equation x^2+((5-2x)/3)^2=2 ?
The solutions to the equation x^2+((5-2x)/3)^2=2 are x=1,x= 7/13
Find the zeros of x^2+((5-2x)/3)^2=2
The zeros of x^2+((5-2x)/3)^2=2 are x=1,x= 7/13