## Algebra 2 (1st Edition)

$150$ and $100$
Our equation is according to the text of the exercise: $3\cdot100\cdot50=(100+x)(50+x)\\15000=5000+x^2+150x\\x^2+150x-10000=0\\(x+200)(x-50)=0$ Thus $x=50$ or $x=−200$ but $x$ has to be positive, thus $x=50$. Thus the new dimensions are $150$ and $100$.