Solve Using the Quadratic Formula (3r+1)^2+(r+2)^2=65

${(3 r + 1)}^{2} + {(r + 2)}^{2} = 65$

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

Subtract $65$ from both sides of the equation.

${(3 r + 1)}^{2} + {(r + 2)}^{2} - 65 = 0$

Simplify ${(3 r + 1)}^{2} + {(r + 2)}^{2} - 65$ .

Simplify each term.

Rewrite ${(3 r + 1)}^{2}$ as $(3 r + 1) (3 r + 1)$ .

$(3 r + 1) (3 r + 1) + {(r + 2)}^{2} - 65 = 0$

Expand $(3 r + 1) (3 r + 1)$ using the FOIL Method.

Apply the distributive property.

$3 r (3 r + 1) + 1 (3 r + 1) + {(r + 2)}^{2} - 65 = 0$

Apply the distributive property.

$3 r (3 r) + 3 r \cdot 1 + 1 (3 r + 1) + {(r + 2)}^{2} - 65 = 0$

Apply the distributive property.

$3 r (3 r) + 3 r \cdot 1 + 1 (3 r) + 1 \cdot 1 + {(r + 2)}^{2} - 65 = 0$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$3 \cdot 3 r \cdot r + 3 r \cdot 1 + 1 (3 r) + 1 \cdot 1 + {(r + 2)}^{2} - 65 = 0$

Multiply $r$ by $r$ by adding the exponents.

Move $r$ .

$3 \cdot 3 (r \cdot r) + 3 r \cdot 1 + 1 (3 r) + 1 \cdot 1 + {(r + 2)}^{2} - 65 = 0$

Multiply $r$ by $r$ .

$3 \cdot 3 r^{2} + 3 r \cdot 1 + 1 (3 r) + 1 \cdot 1 + {(r + 2)}^{2} - 65 = 0$

Multiply $3$ by $3$ .

$9 r^{2} + 3 r \cdot 1 + 1 (3 r) + 1 \cdot 1 + {(r + 2)}^{2} - 65 = 0$

Multiply $3$ by $1$ .

$9 r^{2} + 3 r + 1 (3 r) + 1 \cdot 1 + {(r + 2)}^{2} - 65 = 0$

Multiply $3 r$ by $1$ .

$9 r^{2} + 3 r + 3 r + 1 \cdot 1 + {(r + 2)}^{2} - 65 = 0$

Multiply $1$ by $1$ .

$9 r^{2} + 3 r + 3 r + 1 + {(r + 2)}^{2} - 65 = 0$

Add $3 r$ and $3 r$ .

$9 r^{2} + 6 r + 1 + {(r + 2)}^{2} - 65 = 0$

Rewrite ${(r + 2)}^{2}$ as $(r + 2) (r + 2)$ .

$9 r^{2} + 6 r + 1 + (r + 2) (r + 2) - 65 = 0$

Expand $(r + 2) (r + 2)$ using the FOIL Method.

Apply the distributive property.

$9 r^{2} + 6 r + 1 + r (r + 2) + 2 (r + 2) - 65 = 0$

Apply the distributive property.

$9 r^{2} + 6 r + 1 + r \cdot r + r \cdot 2 + 2 (r + 2) - 65 = 0$

Apply the distributive property.

$9 r^{2} + 6 r + 1 + r \cdot r + r \cdot 2 + 2 r + 2 \cdot 2 - 65 = 0$

Simplify and combine like terms.

Simplify each term.

Multiply $r$ by $r$ .

$9 r^{2} + 6 r + 1 + r^{2} + r \cdot 2 + 2 r + 2 \cdot 2 - 65 = 0$

Move $2$ to the left of $r$ .

$9 r^{2} + 6 r + 1 + r^{2} + 2 \cdot r + 2 r + 2 \cdot 2 - 65 = 0$

Multiply $2$ by $2$ .

$9 r^{2} + 6 r + 1 + r^{2} + 2 r + 2 r + 4 - 65 = 0$

Add $2 r$ and $2 r$ .

$9 r^{2} + 6 r + 1 + r^{2} + 4 r + 4 - 65 = 0$

Add $9 r^{2}$ and $r^{2}$ .

$10 r^{2} + 6 r + 1 + 4 r + 4 - 65 = 0$

Add $6 r$ and $4 r$ .

$10 r^{2} + 10 r + 1 + 4 - 65 = 0$

Add $1$ and $4$ .

$10 r^{2} + 10 r + 5 - 65 = 0$

Subtract $65$ from $5$ .

$10 r^{2} + 10 r - 60 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 10$ , $b = 10$ , and $c = - 60$ into the quadratic formula and solve for $r$ .

$\frac{- 10 \pm \sqrt{10^{2} - 4 \cdot (10 \cdot - 60)}}{2 \cdot 10}$

Simplify.

Simplify the numerator.

Raise $10$ to the power of $2$ .

$r = \frac{- 10 \pm \sqrt{100 - 4 \cdot 10 \cdot - 60}}{2 \cdot 10}$

Multiply $- 4$ by $10$ .

$r = \frac{- 10 \pm \sqrt{100 - 40 \cdot - 60}}{2 \cdot 10}$

Multiply $- 40$ by $- 60$ .

$r = \frac{- 10 \pm \sqrt{100 + 2400}}{2 \cdot 10}$

Add $100$ and $2400$ .

$r = \frac{- 10 \pm \sqrt{2500}}{2 \cdot 10}$

Rewrite $2500$ as $50^{2}$ .

$r = \frac{- 10 \pm \sqrt{50^{2}}}{2 \cdot 10}$

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

$r = \frac{- 10 \pm 50}{2 \cdot 10}$

Multiply $2$ by $10$ .

$r = \frac{- 10 \pm 50}{20}$

Simplify $\frac{- 10 \pm 50}{20}$ .

$r = \frac{- 1 \pm 5}{2}$

The final answer is the combination of both solutions.

$r = 2, - 3$

Solve Using the Quadratic Formula (3r+1)^2+(r+2)^2=65

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