Solve the Equation: Finding X in (x+3)^2 + 2x^2 = 18

To solve the equation $(x+3)^2 + 2x^2 = 18$, we'll follow these steps:

• Step 1: Expand the expression $(x+3)^2$.
• Step 2: Combine and simplify terms to form a standard quadratic equation.
• Step 3: Use the quadratic formula to find values of $x$.

Now, let's work through each step.
Step 1: Expand $(x+3)^2$:

$(x+3)^2 = x^2 + 2 \times x \times 3 + 3^2 = x^2 + 6x + 9$.

Step 2: Substitute back into the original equation:

$x^2 + 6x + 9 + 2x^2 = 18$.

Combine like terms:

$3x^2 + 6x + 9 = 18$.

Subtract 18 from both sides to form the quadratic equation:

$3x^2 + 6x + 9 - 18 = 0$.

Simplify:

$3x^2 + 6x - 9 = 0$.

Divide every term by 3 to simplify further:

$x^2 + 2x - 3 = 0$.

Step 3: Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=1$, $b=2$, and $c=-3$.

Calculate discriminant: $b^2 - 4ac = 2^2 - 4 \times 1 \times (-3) = 4 + 12 = 16$.

Since the discriminant is positive, there are two real roots.

Find roots:

$x = \frac{-2 \pm \sqrt{16}}{2 \times 1}$.

Calculate roots:

$x_1 = \frac{-2 + 4}{2} = 1$,
$x_2 = \frac{-2 - 4}{2} = -3$.

Therefore, the solutions to the equation are $x_1 = 1$, $x_2 = -3$.