## College Algebra (11th Edition)

$x=\left\{ -\dfrac{3}{4},1 \right\}$
$\bf{\text{Solution Outline:}}$ First, factor the given equation, $-4x^2+x=-3 .$ Then equate each factor to zero and solve for the values of the variable. $\bf{\text{Solution Details:}}$ Using the properties of equality, the given equation is equivalent to \begin{array}{l}\require{cancel} -4x^2+x+3=0 \\\\ (-1)(-4x^2+x+3)=(0)(-1) \\\\ 4x^2-x-3=0 .\end{array} The two numbers whose product is $ac= 4(-3)=-12$ and whose sum is $b= -1$ are $\{ -4,3 \}$. Using these two numbers to decompose the middle term of the equation above results to \begin{array}{l}\require{cancel} 4x^2-4x+3x-3=0 \\\\ (4x^2-4x)+(3x-3)=0 \\\\ 4x(x-1)+3(x-1)=0 \\\\ (x-1)(4x+3)=0 .\end{array} Equating each factor to zero (or the Zero-Factor Property), the solutions to the given equation are \begin{array}{l}\require{cancel} x-1=0 \\\\ x=1 ,\\\\\text{OR}\\\\ 4x+3=0 \\\\ 4x=-3 \\\\ x=-\dfrac{3}{4} .\end{array} Hence, the solutions are $x=\left\{ -\dfrac{3}{4},1 \right\} .$