What is the answer to (3x-3)^2-(2x-7)=(3x-5)(3x+5) ?

The answer to (3x-3)^2-(2x-7)=(3x-5)(3x+5) is x= 41/20

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(3x-3)^2-(2x-7)=(3x-5)(3x+5)

(3 x - 3)^{2} - (2 x - 7) = (3 x - 5) (3 x + 5)

x = \frac{41}{20}

Solution steps

(3 x - 3)^{2} - (2 x - 7) = (3 x - 5) (3 x + 5)

Expand

(3 x - 3)^{2} - (2 x - 7) : 9 x^{2} - 20 x + 16

Expand

(3 x - 5) (3 x + 5) : 9 x^{2} - 25

9 x^{2} - 20 x + 16 = 9 x^{2} - 25

Move

16

to the right side

9 x^{2} - 20 x = 9 x^{2} - 41

Move

9 x^{2}

to the left side

- 20 x = - 41

Divide both sides by

- 20

x = \frac{41}{20}

x, 18 x - 7 y = 4 x

- 125 = - 5 k

3 x - 1 + 3 x - 1 = 8 x - 20

3 (a - 4) + 4 = 2 (a - 1)

x + 10 x - 25 = 4 x - 10

What is the answer to (3x-3)^2-(2x-7)=(3x-5)(3x+5) ?
The answer to (3x-3)^2-(2x-7)=(3x-5)(3x+5) is x= 41/20
Solve for x: (3x-3)^2-(2x-7)=(3x-5)(3x+5)
The solution is x= 41/20