## Intermediate Algebra for College Students (7th Edition)

{$\pm 8$}
Given: $2x^2-3=125$ This implies $2x^2-128=0$ Take the common terms out. $2(x^2-64)=0$ This can be written as: $(x^2-8^2)=0$ $(x+8)(x-8)=0$ Thus, $x=\pm 8$ Hence, $x=${$\pm 8$}