## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$x=-5$
Multiplying both sides by $2,$ the given expression is equivalent to \begin{array}{l}\require{cancel} \dfrac{1}{2}x^2+5x+\dfrac{25}{2}=0 \\\\ 2\left( \dfrac{1}{2}x^2+5x+\dfrac{25}{2} \right)=2(0) \\\\ x^2+10x+25=0 .\end{array} Using the factoring of trinomials in the form $x^2+bx+c,$ the $\text{ equation }$ \begin{array}{l}\require{cancel} x^2+10x+25=0 \end{array} has $c= 25$ and $b= 10 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 5,5 \right\}.$ Using these two numbers, the $\text{ equation }$ above is equivalent to \begin{array}{l}\require{cancel} (x+5)(x+5)=0 .\end{array} Equating each factor to zero (Zero Product Property), and then isolating the variable, the solution is $x=-5 .$