Find the Roots (Zeros) (2x+3)^2+(x+4)^2=10

${(2 x + 3)}^{2} + {(x + 4)}^{2} = 10$

Subtract $10$ from both sides of the equation.

${(2 x + 3)}^{2} + {(x + 4)}^{2} - 10 = 0$

Simplify ${(2 x + 3)}^{2} + {(x + 4)}^{2} - 10$ .

Simplify each term.

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

$(2 x + 3) (2 x + 3) + {(x + 4)}^{2} - 10 = 0$

Expand $(2 x + 3) (2 x + 3)$ using the FOIL Method.

Apply the distributive property.

$2 x (2 x + 3) + 3 (2 x + 3) + {(x + 4)}^{2} - 10 = 0$

Apply the distributive property.

$2 x (2 x) + 2 x \cdot 3 + 3 (2 x + 3) + {(x + 4)}^{2} - 10 = 0$

Apply the distributive property.

$2 x (2 x) + 2 x \cdot 3 + 3 (2 x) + 3 \cdot 3 + {(x + 4)}^{2} - 10 = 0$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$2 \cdot 2 x \cdot x + 2 x \cdot 3 + 3 (2 x) + 3 \cdot 3 + {(x + 4)}^{2} - 10 = 0$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$2 \cdot 2 (x \cdot x) + 2 x \cdot 3 + 3 (2 x) + 3 \cdot 3 + {(x + 4)}^{2} - 10 = 0$

Multiply $x$ by $x$ .

$2 \cdot 2 x^{2} + 2 x \cdot 3 + 3 (2 x) + 3 \cdot 3 + {(x + 4)}^{2} - 10 = 0$

Multiply $2$ by $2$ .

$4 x^{2} + 2 x \cdot 3 + 3 (2 x) + 3 \cdot 3 + {(x + 4)}^{2} - 10 = 0$

Multiply $3$ by $2$ .

$4 x^{2} + 6 x + 3 (2 x) + 3 \cdot 3 + {(x + 4)}^{2} - 10 = 0$

Multiply $2$ by $3$ .

$4 x^{2} + 6 x + 6 x + 3 \cdot 3 + {(x + 4)}^{2} - 10 = 0$

Multiply $3$ by $3$ .

$4 x^{2} + 6 x + 6 x + 9 + {(x + 4)}^{2} - 10 = 0$

Add $6 x$ and $6 x$ .

$4 x^{2} + 12 x + 9 + {(x + 4)}^{2} - 10 = 0$

Rewrite ${(x + 4)}^{2}$ as $(x + 4) (x + 4)$ .

$4 x^{2} + 12 x + 9 + (x + 4) (x + 4) - 10 = 0$

Expand $(x + 4) (x + 4)$ using the FOIL Method.

Apply the distributive property.

$4 x^{2} + 12 x + 9 + x (x + 4) + 4 (x + 4) - 10 = 0$

Apply the distributive property.

$4 x^{2} + 12 x + 9 + x \cdot x + x \cdot 4 + 4 (x + 4) - 10 = 0$

Apply the distributive property.

$4 x^{2} + 12 x + 9 + x \cdot x + x \cdot 4 + 4 x + 4 \cdot 4 - 10 = 0$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$4 x^{2} + 12 x + 9 + x^{2} + x \cdot 4 + 4 x + 4 \cdot 4 - 10 = 0$

Move $4$ to the left of $x$ .

$4 x^{2} + 12 x + 9 + x^{2} + 4 \cdot x + 4 x + 4 \cdot 4 - 10 = 0$

Multiply $4$ by $4$ .

$4 x^{2} + 12 x + 9 + x^{2} + 4 x + 4 x + 16 - 10 = 0$

Add $4 x$ and $4 x$ .

$4 x^{2} + 12 x + 9 + x^{2} + 8 x + 16 - 10 = 0$

Add $4 x^{2}$ and $x^{2}$ .

$5 x^{2} + 12 x + 9 + 8 x + 16 - 10 = 0$

Add $12 x$ and $8 x$ .

$5 x^{2} + 20 x + 9 + 16 - 10 = 0$

Add $9$ and $16$ .

$5 x^{2} + 20 x + 25 - 10 = 0$

Subtract $10$ from $25$ .

$5 x^{2} + 20 x + 15 = 0$

Factor the left side of the equation.

Factor $5$ out of $5 x^{2} + 20 x + 15$ .

Factor $5$ out of $5 x^{2}$ .

$5 (x^{2}) + 20 x + 15 = 0$

Factor $5$ out of $20 x$ .

$5 (x^{2}) + 5 (4 x) + 15 = 0$

Factor $5$ out of $15$ .

$5 x^{2} + 5 (4 x) + 5 \cdot 3 = 0$

Factor $5$ out of $5 x^{2} + 5 (4 x)$ .

$5 (x^{2} + 4 x) + 5 \cdot 3 = 0$

Factor $5$ out of $5 (x^{2} + 4 x) + 5 \cdot 3$ .

$5 (x^{2} + 4 x + 3) = 0$

Factor.

Factor $x^{2} + 4 x + 3$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $3$ and whose sum is $4$ .

$1, 3$

Write the factored form using these integers.

$5 ((x + 1) (x + 3)) = 0$

Remove unnecessary parentheses.

$5 (x + 1) (x + 3) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x + 1 = 0$

$x + 3 = 0$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $5 (x + 1) (x + 3) = 0$ true.

$x = - 1, - 3$

Find the Roots (Zeros) (2x+3)^2+(x+4)^2=10

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