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

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

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

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

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

x = - 2, x = - 1

Solution steps

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

Multiply both sides by

3

3 - x^{2} - (3 x + 2) = 3

Expand

3 - x^{2} - (3 x + 2) : - x^{2} - 3 x + 1

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

Move

3

to the left side

- x^{2} - 3 x - 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 1 ) ( - 2 )}{2 ( - 1 )}

(- 3)^{2} - 4 (- 1) (- 2) = 1

x_{1, 2} = \frac{- ( - 3 ) \pm 1}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 1}{2 ( - 1 )}, x_{2} = \frac{- ( - 3 ) - 1}{2 ( - 1 )}

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

x = \frac{- ( - 3 ) - 1}{2 ( - 1 )} : - 1

The solutions to the quadratic equation are:

x = - 2, x = - 1

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

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

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

x y + x + y + 1

x^{2} 150 x^{5} y^{4}

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