Solve Using the Quadratic Formula (2x+3)^2+(x+4)^2=10

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

Step 1

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

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 $25$ .

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

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 $a = 5$ , $b = 20$ , and $c = 15$ into the quadratic formula and solve for $x$ .

$\frac{- 20 \pm \sqrt{20^{2} - 4 \cdot (5 \cdot 15)}}{2 \cdot 5}$

Step 4

Simplify.

Simplify the numerator.

Raise $20$ to the power of $2$ .

$x = \frac{- 20 \pm \sqrt{400 - 4 \cdot 5 \cdot 15}}{2 \cdot 5}$

Multiply $- 4 \cdot 5 \cdot 15$ .

Multiply $- 4$ by $5$ .

$x = \frac{- 20 \pm \sqrt{400 - 20 \cdot 15}}{2 \cdot 5}$

Multiply $- 20$ by $15$ .

$x = \frac{- 20 \pm \sqrt{400 - 300}}{2 \cdot 5}$

Subtract $300$ from $400$ .

$x = \frac{- 20 \pm \sqrt{100}}{2 \cdot 5}$

Rewrite $100$ as $10^{2}$ .

$x = \frac{- 20 \pm \sqrt{10^{2}}}{2 \cdot 5}$

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

$x = \frac{- 20 \pm 10}{2 \cdot 5}$

Multiply $2$ by $5$ .

$x = \frac{- 20 \pm 10}{10}$

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

$x = - 2 \pm 1$

Step 5

The final answer is the combination of both solutions.

$x = - 1, - 3$

Solve Using the Quadratic Formula (2x+3)^2+(x+4)^2=10

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