Solve Using the Quadratic Formula (t+2)^2+(t-3)^2=9

${(t + 2)}^{2} + {(t - 3)}^{2} = 9$

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

Subtract $9$ from both sides of the equation.

${(t + 2)}^{2} + {(t - 3)}^{2} - 9 = 0$

Simplify ${(t + 2)}^{2} + {(t - 3)}^{2} - 9$ .

Simplify each term.

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

$(t + 2) (t + 2) + {(t - 3)}^{2} - 9 = 0$

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

Apply the distributive property.

$t (t + 2) + 2 (t + 2) + {(t - 3)}^{2} - 9 = 0$

Apply the distributive property.

$t \cdot t + t \cdot 2 + 2 (t + 2) + {(t - 3)}^{2} - 9 = 0$

Apply the distributive property.

$t \cdot t + t \cdot 2 + 2 t + 2 \cdot 2 + {(t - 3)}^{2} - 9 = 0$

Simplify and combine like terms.

Simplify each term.

Multiply $t$ by $t$ .

$t^{2} + t \cdot 2 + 2 t + 2 \cdot 2 + {(t - 3)}^{2} - 9 = 0$

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

$t^{2} + 2 \cdot t + 2 t + 2 \cdot 2 + {(t - 3)}^{2} - 9 = 0$

Multiply $2$ by $2$ .

$t^{2} + 2 t + 2 t + 4 + {(t - 3)}^{2} - 9 = 0$

Add $2 t$ and $2 t$ .

$t^{2} + 4 t + 4 + {(t - 3)}^{2} - 9 = 0$

Rewrite ${(t - 3)}^{2}$ as $(t - 3) (t - 3)$ .

$t^{2} + 4 t + 4 + (t - 3) (t - 3) - 9 = 0$

Expand $(t - 3) (t - 3)$ using the FOIL Method.

Apply the distributive property.

$t^{2} + 4 t + 4 + t (t - 3) - 3 (t - 3) - 9 = 0$

Apply the distributive property.

$t^{2} + 4 t + 4 + t \cdot t + t \cdot - 3 - 3 (t - 3) - 9 = 0$

Apply the distributive property.

$t^{2} + 4 t + 4 + t \cdot t + t \cdot - 3 - 3 t - 3 \cdot - 3 - 9 = 0$

Simplify and combine like terms.

Simplify each term.

Multiply $t$ by $t$ .

$t^{2} + 4 t + 4 + t^{2} + t \cdot - 3 - 3 t - 3 \cdot - 3 - 9 = 0$

Move $- 3$ to the left of $t$ .

$t^{2} + 4 t + 4 + t^{2} - 3 \cdot t - 3 t - 3 \cdot - 3 - 9 = 0$

Multiply $- 3$ by $- 3$ .

$t^{2} + 4 t + 4 + t^{2} - 3 t - 3 t + 9 - 9 = 0$

Subtract $3 t$ from $- 3 t$ .

$t^{2} + 4 t + 4 + t^{2} - 6 t + 9 - 9 = 0$

Combine the opposite terms in $t^{2} + 4 t + 4 + t^{2} - 6 t + 9 - 9$ .

Subtract $9$ from $9$ .

$t^{2} + 4 t + 4 + t^{2} - 6 t + 0 = 0$

Add $t^{2} + 4 t + 4 + t^{2} - 6 t$ and $0$ .

$t^{2} + 4 t + 4 + t^{2} - 6 t = 0$

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

$2 t^{2} + 4 t + 4 - 6 t = 0$

Subtract $6 t$ from $4 t$ .

$2 t^{2} - 2 t + 4 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 2$ , $b = - 2$ , and $c = 4$ into the quadratic formula and solve for $t$ .

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

Simplify.

Simplify the numerator.

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

$t = \frac{2 \pm \sqrt{4 - 4 \cdot 2 \cdot 4}}{2 \cdot 2}$

Multiply $- 4$ by $2$ .

$t = \frac{2 \pm \sqrt{4 - 8 \cdot 4}}{2 \cdot 2}$

Multiply $- 8$ by $4$ .

$t = \frac{2 \pm \sqrt{4 - 32}}{2 \cdot 2}$

Subtract $32$ from $4$ .

$t = \frac{2 \pm \sqrt{- 28}}{2 \cdot 2}$

Rewrite $- 28$ as $- 1 (28)$ .

$t = \frac{2 \pm \sqrt{- 1 \cdot 28}}{2 \cdot 2}$

Rewrite $\sqrt{- 1 (28)}$ as $\sqrt{- 1} \cdot \sqrt{28}$ .

$t = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{28}}{2 \cdot 2}$

Rewrite $\sqrt{- 1}$ as $i$ .

$t = \frac{2 \pm i \cdot \sqrt{28}}{2 \cdot 2}$

Rewrite $28$ as $2^{2} \cdot 7$ .

Factor $4$ out of $28$ .

$t = \frac{2 \pm i \cdot \sqrt{4 (7)}}{2 \cdot 2}$

Rewrite $4$ as $2^{2}$ .

$t = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 7}}{2 \cdot 2}$

Pull terms out from under the radical.

$t = \frac{2 \pm i \cdot (2 \sqrt{7})}{2 \cdot 2}$

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

$t = \frac{2 \pm 2 i \sqrt{7}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$t = \frac{2 \pm 2 i \sqrt{7}}{4}$

Simplify $\frac{2 \pm 2 i \sqrt{7}}{4}$ .

$t = \frac{1 \pm i \sqrt{7}}{2}$

Solve Using the Quadratic Formula (t+2)^2+(t-3)^2=9

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