Solve Using the Quadratic Formula (x+5)^2+(x-2)^2=37

${(x + 5)}^{2} + {(x - 2)}^{2} = 37$

Step 1

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

Subtract $37$ from both sides of the equation.

${(x + 5)}^{2} + {(x - 2)}^{2} - 37 = 0$

Simplify ${(x + 5)}^{2} + {(x - 2)}^{2} - 37$ .

Simplify each term.

Rewrite ${(x + 5)}^{2}$ as $(x + 5) (x + 5)$ .

$(x + 5) (x + 5) + {(x - 2)}^{2} - 37 = 0$

Expand $(x + 5) (x + 5)$ using the FOIL Method.

Apply the distributive property.

$x (x + 5) + 5 (x + 5) + {(x - 2)}^{2} - 37 = 0$

Apply the distributive property.

$x \cdot x + x \cdot 5 + 5 (x + 5) + {(x - 2)}^{2} - 37 = 0$

Apply the distributive property.

$x \cdot x + x \cdot 5 + 5 x + 5 \cdot 5 + {(x - 2)}^{2} - 37 = 0$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + x \cdot 5 + 5 x + 5 \cdot 5 + {(x - 2)}^{2} - 37 = 0$

Move $5$ to the left of $x$ .

$x^{2} + 5 \cdot x + 5 x + 5 \cdot 5 + {(x - 2)}^{2} - 37 = 0$

Multiply $5$ by $5$ .

$x^{2} + 5 x + 5 x + 25 + {(x - 2)}^{2} - 37 = 0$

Add $5 x$ and $5 x$ .

$x^{2} + 10 x + 25 + {(x - 2)}^{2} - 37 = 0$

Rewrite ${(x - 2)}^{2}$ as $(x - 2) (x - 2)$ .

$x^{2} + 10 x + 25 + (x - 2) (x - 2) - 37 = 0$

Expand $(x - 2) (x - 2)$ using the FOIL Method.

Apply the distributive property.

$x^{2} + 10 x + 25 + x (x - 2) - 2 (x - 2) - 37 = 0$

Apply the distributive property.

$x^{2} + 10 x + 25 + x \cdot x + x \cdot - 2 - 2 (x - 2) - 37 = 0$

Apply the distributive property.

$x^{2} + 10 x + 25 + x \cdot x + x \cdot - 2 - 2 x - 2 \cdot - 2 - 37 = 0$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + 10 x + 25 + x^{2} + x \cdot - 2 - 2 x - 2 \cdot - 2 - 37 = 0$

Move $- 2$ to the left of $x$ .

$x^{2} + 10 x + 25 + x^{2} - 2 \cdot x - 2 x - 2 \cdot - 2 - 37 = 0$

Multiply $- 2$ by $- 2$ .

$x^{2} + 10 x + 25 + x^{2} - 2 x - 2 x + 4 - 37 = 0$

Subtract $2 x$ from $- 2 x$ .

$x^{2} + 10 x + 25 + x^{2} - 4 x + 4 - 37 = 0$

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

$2 x^{2} + 10 x + 25 - 4 x + 4 - 37 = 0$

Subtract $4 x$ from $10 x$ .

$2 x^{2} + 6 x + 25 + 4 - 37 = 0$

Add $25$ and $4$ .

$2 x^{2} + 6 x + 29 - 37 = 0$

Subtract $37$ from $29$ .

$2 x^{2} + 6 x - 8 = 0$

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 $c = - 8$ into the quadratic formula and solve for $x$ .

$\frac{- 6 \pm \sqrt{6^{2} - 4 \cdot (2 \cdot - 8)}}{2 \cdot 2}$

Step 4

Simplify.

Simplify the numerator.

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

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 2 \cdot - 8}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 8$ .

Multiply $- 4$ by $2$ .

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

Multiply $- 8$ by $- 8$ .

$x = \frac{- 6 \pm \sqrt{36 + 64}}{2 \cdot 2}$

Add $36$ and $64$ .

$x = \frac{- 6 \pm \sqrt{100}}{2 \cdot 2}$

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

$x = \frac{- 6 \pm \sqrt{10^{2}}}{2 \cdot 2}$

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

$x = \frac{- 6 \pm 10}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 6 \pm 10}{4}$

Simplify $\frac{- 6 \pm 10}{4}$ .

$x = \frac{- 3 \pm 5}{2}$

Step 5

The final answer is the combination of both solutions.

$x = 1, - 4$

Solve Using the Quadratic Formula (x+5)^2+(x-2)^2=37

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