## Precalculus (6th Edition)

$\color{blue}{x=-1, x=3}$
Square both sides to obtain: $4x+13=(2x-1)^2 \\4x+13=(2x)^2-2(2x)(1)+(-1)^2 \\4x+13=4x^2-4x+1$ Move all terms to the left side of the equation. Note that when a term is transferred to the other side of the equation, its sign changes to its opposite. $0=4x^2-4x+1-4x-13 \\0=4x^2-8x-12 \\4x^2-8x-12=0$ Factor out $4$ to obtain: $4(x^2-2x-3)=0$ Factor the trinomial to obtain: $4(x-3)(x+1)=0$ Divide $4$ to both sides to obtain: $(x-3)(x+1)=0$ Use the zero-factor property by equating each factor to zero. $x-3=0$ or $x+1=0$ Solve each equation to obtain: $x=3$ or $x=-1$. Upon checking, both proposed solutions satisfy the original equation. Thus, the solutions to the given equation are $\color{blue}{x=-1, x=3}$.