What are the solutions to the equation x+(3x+1)/2-(x-2)/3 =x^2-2 ?

The solutions to the equation x+(3x+1)/2-(x-2)/3 =x^2-2 are x= 19/6 ,x=-1

Find the zeros of x+(3x+1)/2-(x-2)/3 =x^2-2

The zeros of x+(3x+1)/2-(x-2)/3 =x^2-2 are x= 19/6 ,x=-1

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

x = \frac{19}{6}, x = - 1

Decimal Notation

x = 3.16666 \dots, x = - 1

Solution steps

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

Find Least Common Multiplier of

2, 3 : 6

Multiply by LCM=

6

x \cdot 6 + \frac{3 x + 1}{2} \cdot 6 - \frac{x - 2}{3} \cdot 6 = x^{2} \cdot 6 - 2 \cdot 6

Simplify

6 x + 3 (3 x + 1) - 2 (x - 2) = 6 x^{2} - 12

Expand

6 x + 3 (3 x + 1) - 2 (x - 2) : 13 x + 7

13 x + 7 = 6 x^{2} - 12

Switch sides

6 x^{2} - 12 = 13 x + 7

Move

7

to the left side

6 x^{2} - 19 = 13 x

Move

13 x

to the left side

6 x^{2} - 19 - 13 x = 0

Write in the standard form

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

6 x^{2} - 13 x - 19 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 6 ( - 19 )}{2 \cdot 6}

(- 13)^{2} - 4 \cdot 6 (- 19) = 25

x_{1, 2} = \frac{- ( - 13 ) \pm 25}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 13 ) + 25}{2 \cdot 6}, x_{2} = \frac{- ( - 13 ) - 25}{2 \cdot 6}

x = \frac{- ( - 13 ) + 25}{2 \cdot 6} : \frac{19}{6}

x = \frac{- ( - 13 ) - 25}{2 \cdot 6} : - 1

The solutions to the quadratic equation are:

x = \frac{19}{6}, x = - 1

d (8 y - 1) - (1 - 8 y)

\frac{3 x ^{- 2}}{y}

2 x^{2} = - 1 - 6 x

\frac{3}{4} y (\frac{a}{2} - \frac{b}{3}) - \frac{2}{3} x (\frac{b}{3} - \frac{a}{2})

\frac{7}{4} - 1

What are the solutions to the equation x+(3x+1)/2-(x-2)/3 =x^2-2 ?
The solutions to the equation x+(3x+1)/2-(x-2)/3 =x^2-2 are x= 19/6 ,x=-1
Find the zeros of x+(3x+1)/2-(x-2)/3 =x^2-2
The zeros of x+(3x+1)/2-(x-2)/3 =x^2-2 are x= 19/6 ,x=-1