What are the solutions to the equation ((x-2)^2)/2+(5x+6)/6 =((x+3)(x-3))/3+6 ?

The solutions to the equation ((x-2)^2)/2+(5x+6)/6 =((x+3)(x-3))/3+6 are x=7,x=0

Find the zeros of ((x-2)^2)/2+(5x+6)/6 =((x+3)(x-3))/3+6

The zeros of ((x-2)^2)/2+(5x+6)/6 =((x+3)(x-3))/3+6 are x=7,x=0

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\frac{( x - 2 ) ^{2}}{2} + \frac{5 x + 6}{6} = \frac{( x + 3 ) ( x - 3 )}{3} + 6

x = 7, x = 0

Solution steps

\frac{( x - 2 ) ^{2}}{2} + \frac{5 x + 6}{6} = \frac{( x + 3 ) ( x - 3 )}{3} + 6

Find Least Common Multiplier of

2, 6, 3 : 6

Multiply by LCM=

6

\frac{( x - 2 ) ^{2}}{2} \cdot 6 + \frac{5 x + 6}{6} \cdot 6 = \frac{( x + 3 ) ( x - 3 )}{3} \cdot 6 + 6 \cdot 6

Simplify

3 (x - 2)^{2} + 5 x + 6 = 2 (x + 3) (x - 3) + 36

Expand

3 (x - 2)^{2} + 5 x + 6 : 3 x^{2} - 7 x + 18

Expand

2 (x + 3) (x - 3) + 36 : 2 x^{2} + 18

3 x^{2} - 7 x + 18 = 2 x^{2} + 18

Move

18

to the left side

3 x^{2} - 7 x = 2 x^{2}

Move

2 x^{2}

to the left side

x^{2} - 7 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 7)^{2} - 4 \cdot 1 \cdot 0 = 7

x_{1, 2} = \frac{- ( - 7 ) \pm 7}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 7}{2 \cdot 1}, x_{2} = \frac{- ( - 7 ) - 7}{2 \cdot 1}

x = \frac{- ( - 7 ) + 7}{2 \cdot 1} : 7

x = \frac{- ( - 7 ) - 7}{2 \cdot 1} : 0

The solutions to the quadratic equation are:

x = 7, x = 0

z^{2} + 9 = 10 z

(\frac{6}{11})^{2}

\frac{4 x ^{4} y ^{2} z}{x ^{2} yz}

- 10 - 5 n^{2} = - 330

\frac{x}{- 4} \geq 8

What are the solutions to the equation ((x-2)^2)/2+(5x+6)/6 =((x+3)(x-3))/3+6 ?
The solutions to the equation ((x-2)^2)/2+(5x+6)/6 =((x+3)(x-3))/3+6 are x=7,x=0
Find the zeros of ((x-2)^2)/2+(5x+6)/6 =((x+3)(x-3))/3+6
The zeros of ((x-2)^2)/2+(5x+6)/6 =((x+3)(x-3))/3+6 are x=7,x=0