Question

(x+3)^2-(x-4)(x-2)=0

Answer 1

(x+3)^2 - (x-4)(x-2) = 0

x^2 + 6x + 9 - (x^2 - 6x + 8) = 0

x^2 + 6x + 9 - x^2 + 6x - 8 = 0

12x = -1

x = -1/12

Ответ: -1/12

Answer 2

To solve the equation $(x+3)^2 - (x-4)(x-2) = 0$, we'll start by simplifying the left-hand side of the equation and then proceed with solving for the value(s) of $x$.

Step 1: Expand the expressions. $(x+3)^2 = (x+3)(x+3) = x^2 + 6x + 9$ $(x-4)(x-2) = x^2 - 2x - 4x + 8 = x^2 - 6x + 8$

Step 2: Substitute the expanded expressions back into the equation. $x^2 + 6x + 9 - (x^2 - 6x + 8) = 0$

Step 3: Simplify the equation. $x^2 + 6x + 9 - x^2 + 6x - 8 = 0$ $12x + 1 = 0$

Step 4: Solve for $x$. $12x = -1$ $x = -1/12$

So, the solution to the equation is $x = -\frac{1}{12}$.