Properties of zero word problem..stuck

[xstellar1x]

Please help...getting stuck....

A rectangular parking lot is 30 ft longer than its width. Find the dimensions of the parking lot if it measures 150 ft diagonally..
This is what I have so far:

x^2+(x+30)^2= (150)^2
x^2 + x^2 + 60x + 1200 = 22500
2x^2+60x-2130=0

Here is where I get stuck.....I am not good at factoring...I am working on it....slowly

Thanks for the help!

 

[BigGlenntheHeavy]

\(\displaystyle Your \ final \ answer \ before \ factoring \ should \ be:\)

\(\displaystyle x^2+30x-10800 \ = \ 0, \ now \ if \ factoring \ is \ a \ hasstle, \ plug \ in \ the \ quadratic \ formula, \ to \ wit:\)

\(\displaystyle x \ = \ \frac{-30 \ \pm \sqrt{30^2+4(10800)}}{2} \ = \ \frac{-30 \ \pm210}{2} \ = \ \frac{180}{2} \ = \ 90.\)

\(\displaystyle The \ rest \ should \ be \ elementary.\)

 

[Denis]

\(\displaystyle \ now \ if \ factoring \ is \ a \ hasstle,\)

HASSLE. Go stand in the corner!

 

[xstellar1x]

I am sorry....I am older and taking this course online so I get lost easily....How did you get :

X^2 + 30x - 10800=0 I guess I am not seeing how you got 10800? Am i just dense?? :?

 

[masters]

I am sorry....I am older and taking this course online so I get lost easily....How did you get :

X^2 + 30x - 10800=0 I guess I am not seeing how you got 10800? Am i just dense?? :?

Hi xstellar1x,

The second line of your attempt was incorrect. BigG corrected it for you, but he spared the details.

\(\displaystyle x^2+(x+30)^2=(150)^2\)

\(\displaystyle x^2+x^2+60x+900=22500\)

\(\displaystyle 2x^2+60x-21600=0\)

Divide everything by 2

\(\displaystyle x^2+30x-10800=0\)

 

[lookagain]

\(\displaystyle \ \frac{-30 \ \pm210}{2} \ = \ \frac{180}{2} \ = \ 90.\)

xstellar1x,

instead of what's in the quote box here, it should be closer to this:

\(\displaystyle x = \frac{-30\ \pm 210}{2} \ becomes\)

\(\displaystyle x = \ \frac {-30 - 210}{2} = \frac{-240}{2} \ = \ -120 \ \\)

\(\displaystyle \ \ \ \ We \ \ discard \ \ this \ \ because \ \ the \ \ sides \ \ must \ \ have \ \ positive \ \ lengths.\)


\(\displaystyle OR \ \\) \(\displaystyle . . . \ \ x = \ \frac {-30 + 210}{2} \ = \ \frac{180}{2} \ = \ 90\)

 

[xstellar1x]

Thank you all so much!!!