Question

(3х+1)^2-(3х-2)(2+3х=17

Answer 1

$(3x+1)^2-(3x-2)(2+3x)=17\\\\9x^2+6x+1-(9x^2-4)=17\\\\9x^2+6x+1-9x^2+4=17\\\\6x+5=17\\\\6x=12\\\\\boxed{x=2}$

Answer 2

It seems there's a typo in the equation you provided. The equal sign "=" appears twice, which makes it ambiguous. I'll assume you meant to write:

(3x + 1)^2 - (3x - 2)(2 + 3x) = 17

To solve this equation, we'll first simplify the expressions on both sides and then proceed with finding the value of x.

Step 1: Expand the expressions (3x + 1)^2 = (3x + 1)(3x + 1) = 9x^2 + 6x + 1

(3x - 2)(2 + 3x) = 3x * 2 + 3x * 3x - 2 * 2 - 2 * 3x = 6x + 9x^2 - 4 - 6x = 9x^2 - 4

Now the equation becomes:

(9x^2 + 6x + 1) - (9x^2 - 4) = 17

Step 2: Combine like terms

9x^2 + 6x + 1 - 9x^2 + 4 = 17

Step 3: Simplify

6x + 5 = 17

Step 4: Isolate x

Subtract 5 from both sides:

6x = 17 - 5 6x = 12

Now, divide by 6 to find the value of x:

x = 12 / 6 x = 2

So, the solution to the equation is x = 2.