Solve for X in the Quadratic Equation: (x-5)²-5=10+2x

To solve the given equation $(x-5)^2 - 5 = 10 + 2x$, we'll follow these steps:

• Step 1: Expand and simplify the left side of the equation.
• Step 2: Move all terms to form a standard quadratic equation.
• Step 3: Use the quadratic formula to find the values of $x$.

Now, let's work through each step:

Step 1: Expand the left side.
$(x-5)^2 = x^2 - 10x + 25$
The equation becomes:
$x^2 - 10x + 25 - 5 = 10 + 2x$

Step 2: Collect all terms on one side.
$x^2 - 10x + 20 = 10 + 2x$
Subtract $10 + 2x$ from both sides to get:
$x^2 - 10x + 20 - 10 - 2x = 0$
This simplifies to:
$x^2 - 12x + 10 = 0$

Step 3: Apply the quadratic formula:
For $ax^2 + bx + c = 0$, the formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
Here, $a = 1$, $b = -12$, $c = 10$.
Calculate the discriminant:
$b^2 - 4ac = (-12)^2 - 4 \cdot 1 \cdot 10 = 144 - 40 = 104$
Now, solve for $x$:
$x = \frac{-(-12) \pm \sqrt{104}}{2 \cdot 1} = \frac{12 \pm \sqrt{104}}{2}$

Therefore, the solutions to the equation are:
$x_1 = 6 + \frac{\sqrt{104}}{2}$, $x_2 = 6 - \frac{\sqrt{104}}{2}$.

This matches the correct choice, confirming that the solution is correct.