What are the solutions to the equation (x^2-6)/3-(2x^2-2)/2 =2+(x-12)/3 ?

The solutions to the equation (x^2-6)/3-(2x^2-2)/2 =2+(x-12)/3 are x=-3/2 ,x=1

Find the zeros of (x^2-6)/3-(2x^2-2)/2 =2+(x-12)/3

The zeros of (x^2-6)/3-(2x^2-2)/2 =2+(x-12)/3 are x=-3/2 ,x=1

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\frac{x ^{2} - 6}{3} - \frac{2 x ^{2} - 2}{2} = 2 + \frac{x - 12}{3}

x = - \frac{3}{2}, x = 1

Decimal Notation

x = - 1.5, x = 1

Solution steps

\frac{x ^{2} - 6}{3} - \frac{2 x ^{2} - 2}{2} = 2 + \frac{x - 12}{3}

Find Least Common Multiplier of

3, 2 : 6

Multiply by LCM=

6

\frac{x ^{2} - 6}{3} \cdot 6 - \frac{2 x ^{2} - 2}{2} \cdot 6 = 2 \cdot 6 + \frac{x - 12}{3} \cdot 6

Simplify

2 (x^{2} - 6) - 6 (x^{2} - 1) = 12 + 2 (x - 12)

Expand

2 (x^{2} - 6) - 6 (x^{2} - 1) : - 4 x^{2} - 6

Expand

12 + 2 (x - 12) : 2 x - 12

- 4 x^{2} - 6 = 2 x - 12

Move

12

to the left side

- 4 x^{2} + 6 = 2 x

Move

2 x

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 4 ) \cdot 6}{2 ( - 4 )}

(- 2)^{2} - 4 (- 4) \cdot 6 = 10

x_{1, 2} = \frac{- ( - 2 ) \pm 10}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 10}{2 ( - 4 )}, x_{2} = \frac{- ( - 2 ) - 10}{2 ( - 4 )}

x = \frac{- ( - 2 ) + 10}{2 ( - 4 )} : - \frac{3}{2}

x = \frac{- ( - 2 ) - 10}{2 ( - 4 )} : 1

The solutions to the quadratic equation are:

x = - \frac{3}{2}, x = 1

1 - \frac{x ^{2}}{3} - \frac{3 x + 2}{3} = 1

n^{3} + 3 n^{2} + 2 n

\frac{( n + 3 ) !}{n !}

\frac{2}{a} - \frac{3}{a ^{2}}

x y + x + y + 1

What are the solutions to the equation (x^2-6)/3-(2x^2-2)/2 =2+(x-12)/3 ?
The solutions to the equation (x^2-6)/3-(2x^2-2)/2 =2+(x-12)/3 are x=-3/2 ,x=1
Find the zeros of (x^2-6)/3-(2x^2-2)/2 =2+(x-12)/3
The zeros of (x^2-6)/3-(2x^2-2)/2 =2+(x-12)/3 are x=-3/2 ,x=1