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

The solutions to the equation 3x-(2-x)-2=(x+1)(2-3x) are x=-(5+sqrt(97))/6 ,x=(sqrt(97)-5)/6

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

The zeros of 3x-(2-x)-2=(x+1)(2-3x) are x=-(5+sqrt(97))/6 ,x=(sqrt(97)-5)/6

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3 x - (2 - x) - 2 = (x + 1) (2 - 3 x)

x = - \frac{5 + 97}{6}, x = \frac{97 - 5}{6}

Solution steps

3 x - (2 - x) - 2 = (x + 1) (2 - 3 x)

Expand

3 x - (2 - x) - 2 : 4 x - 4

Expand

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

4 x - 4 = - 3 x^{2} - x + 2

Switch sides

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

Move

4

to the left side

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

Move

4 x

to the left side

- 3 x^{2} - 5 x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 3 ) \cdot 6}{2 ( - 3 )}

(- 5)^{2} - 4 (- 3) \cdot 6 = 97

x_{1, 2} = \frac{- ( - 5 ) \pm 97}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 97}{2 ( - 3 )}, x_{2} = \frac{- ( - 5 ) - 97}{2 ( - 3 )}

x = \frac{- ( - 5 ) + 97}{2 ( - 3 )} : - \frac{5 + 97}{6}

x = \frac{- ( - 5 ) - 97}{2 ( - 3 )} : \frac{97 - 5}{6}

The solutions to the quadratic equation are:

x = - \frac{5 + 97}{6}, x = \frac{97 - 5}{6}

x^{2} - 7 x - 10 = - 2 x - 4

x, 4 x^{2} - 4 y^{2} = 0

3 r^{2} + 3 r - 4 = 0

- 5 (2 x - 3)^{2} = - 125

x, x^{2} - 8 k x - 9 = 0

What are the solutions to the equation 3x-(2-x)-2=(x+1)(2-3x) ?
The solutions to the equation 3x-(2-x)-2=(x+1)(2-3x) are x=-(5+sqrt(97))/6 ,x=(sqrt(97)-5)/6
Find the zeros of 3x-(2-x)-2=(x+1)(2-3x)
The zeros of 3x-(2-x)-2=(x+1)(2-3x) are x=-(5+sqrt(97))/6 ,x=(sqrt(97)-5)/6