What is the answer to (3x-1)^2-5(x-2)-(2x+3)^2-(5x+2)(x-1)=0 ?

The answer to (3x-1)^2-5(x-2)-(2x+3)^2-(5x+2)(x-1)=0 is x= 1/5

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(3x-1)^2-5(x-2)-(2x+3)^2-(5x+2)(x-1)=0

(3 x - 1)^{2} - 5 (x - 2) - (2 x + 3)^{2} - (5 x + 2) (x - 1) = 0

x = \frac{1}{5}

Solution steps

(3 x - 1)^{2} - 5 (x - 2) - (2 x + 3)^{2} - (5 x + 2) (x - 1) = 0

Expand

(3 x - 1)^{2} - 5 (x - 2) - (2 x + 3)^{2} - (5 x + 2) (x - 1) : - 20 x + 4

- 20 x + 4 = 0

Move

4

to the right side

- 20 x = - 4

Divide both sides by

- 20

x = \frac{1}{5}

9 r^{2} - 3 = - 152

- \frac{\frac{4}{5}}{\frac{3}{5}}

7 x - 4 = 3 + 8 x

3 (6)^{2}

2 (7 x - 10) = 9 x - 32

What is the answer to (3x-1)^2-5(x-2)-(2x+3)^2-(5x+2)(x-1)=0 ?
The answer to (3x-1)^2-5(x-2)-(2x+3)^2-(5x+2)(x-1)=0 is x= 1/5
Solve for x: (3x-1)^2-5(x-2)-(2x+3)^2-(5x+2)(x-1)=0
The solution is x= 1/5