Chapter 4 Quadratic Functions and Factoring - 4.3 Solve x(squared) + bx + c = 0 - Guided Practice for Examples 3 and 4 - Page 254: 9

$1200$ and $500$.

Our equation is: $2\cdot1000\cdot300=(1000+x)(300+x)\\600000=300000+1300x+x^2\\x^2+1300x-300000=0\\(x+1500)(x-200)=0$ Thus $x=-1500$ or $x=200$ but $x$ cannot be negative, thus the new dimensions will be $1200$ and $500$.