Solve Using the Quadratic Formula x^2+(x+3)^2=15^2

$x^{2} + {(x + 3)}^{2} = 15^{2}$

Step 1

Move all terms to the left side of the equation and simplify.

Simplify the right side.

Raise $15$ to the power of $2$ .

$x^{2} + {(x + 3)}^{2} = 225$

Subtract $225$ from both sides of the equation.

$x^{2} + {(x + 3)}^{2} - 225 = 0$

Simplify $x^{2} + {(x + 3)}^{2} - 225$ .

Simplify each term.

Rewrite ${(x + 3)}^{2}$ as $(x + 3) (x + 3)$ .

$x^{2} + (x + 3) (x + 3) - 225 = 0$

Expand $(x + 3) (x + 3)$ using the FOIL Method.

Apply the distributive property.

$x^{2} + x (x + 3) + 3 (x + 3) - 225 = 0$

Apply the distributive property.

$x^{2} + x \cdot x + x \cdot 3 + 3 (x + 3) - 225 = 0$

Apply the distributive property.

$x^{2} + x \cdot x + x \cdot 3 + 3 x + 3 \cdot 3 - 225 = 0$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + x^{2} + x \cdot 3 + 3 x + 3 \cdot 3 - 225 = 0$

Move $3$ to the left of $x$ .

$x^{2} + x^{2} + 3 \cdot x + 3 x + 3 \cdot 3 - 225 = 0$

Multiply $3$ by $3$ .

$x^{2} + x^{2} + 3 x + 3 x + 9 - 225 = 0$

Add $3 x$ and $3 x$ .

$x^{2} + x^{2} + 6 x + 9 - 225 = 0$

Add $x^{2}$ and $x^{2}$ .

$2 x^{2} + 6 x + 9 - 225 = 0$

Subtract $225$ from $9$ .

$2 x^{2} + 6 x - 216 = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 2$ , $b = 6$ , and $c = - 216$ into the quadratic formula and solve for $x$ .

$\frac{- 6 \pm \sqrt{6^{2} - 4 \cdot (2 \cdot - 216)}}{2 \cdot 2}$

Step 4

Simplify.

Simplify the numerator.

Raise $6$ to the power of $2$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 2 \cdot - 216}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 216$ .

Multiply $- 4$ by $2$ .

$x = \frac{- 6 \pm \sqrt{36 - 8 \cdot - 216}}{2 \cdot 2}$

Multiply $- 8$ by $- 216$ .

$x = \frac{- 6 \pm \sqrt{36 + 1728}}{2 \cdot 2}$

Add $36$ and $1728$ .

$x = \frac{- 6 \pm \sqrt{1764}}{2 \cdot 2}$

Rewrite $1764$ as $42^{2}$ .

$x = \frac{- 6 \pm \sqrt{42^{2}}}{2 \cdot 2}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{- 6 \pm 42}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 6 \pm 42}{4}$

Simplify $\frac{- 6 \pm 42}{4}$ .

$x = \frac{- 3 \pm 21}{2}$

Step 5

The final answer is the combination of both solutions.

$x = 9, - 12$

Solve Using the Quadratic Formula x^2+(x+3)^2=15^2

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