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

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

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

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

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

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

Decimal Notation

x = - 1, x = 0.16666 \dots

Solution steps

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

Expand

(3 x + 1) (1 - x) - 3 (x + 1)^{2} : - 6 x^{2} - 4 x - 2

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

Move

3

to the left side

- 6 x^{2} - 4 x + 1 = x

Move

x

to the left side

- 6 x^{2} - 5 x + 1 = 0

Solve with the quadratic formula

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

(- 5)^{2} - 4 (- 6) \cdot 1 = 7

x_{1, 2} = \frac{- ( - 5 ) \pm 7}{2 ( - 6 )}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 7}{2 ( - 6 )}, x_{2} = \frac{- ( - 5 ) - 7}{2 ( - 6 )}

x = \frac{- ( - 5 ) + 7}{2 ( - 6 )} : - 1

x = \frac{- ( - 5 ) - 7}{2 ( - 6 )} : \frac{1}{6}

The solutions to the quadratic equation are:

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

6 x^{2} + 10 x - 5 = 0

(x^{2} + 15) = 0

(x - 3)^{2} - 1 = - \frac{3}{4}

12 x^{2} - 20 x + 5 = 0

5 x^{2} = 1 - 8 x

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