# Chapter P - Section P.7 - Equations - Exercise Set - Page 106: 70

$\left\{-6, 2\right\}$

#### Work Step by Step

RECALL: To solve $x^2 + bx=c$ by completing the square, add $\left(\frac{b}{2}\right)^2$ on both sides of the equation. Thus, to solve the given equation by completing the square, add $(\frac{4}{2})^2=2^2=4$ on both sides of the equation to obtain: $x^2+4x+4=12+4 \\x^2+4x+4=16 \\(x+2)^2=16$ Take the square root of both sides to obtain: $x+2=\pm \sqrt{16} \\x+2=\pm \sqrt{4^2} \\x+2=\pm 4$ Adad $-2$ on both sides of the equation to obtain: $x = -2 \pm 4$ We get two solutions: $x_1 = -2+4=2 \\x_2 = -2-4=-6$ Therefore, the solution of the given equation is: $\left\{-6, 2\right\}$.

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