# Thread: Unsure of how to solve this right triangle

1. ## Unsure of how to solve this right triangle

Thats 29 squared on the hypotenuse and x+3 on the opposite and just x on the adjacent. I cant figure what to do with the 29 after it is given an exponent of 2 and taken out of square form.

2. ## Re: Unsure of how to solve this right triangle

Originally Posted by goldbug78

Thats 29 squared on the hypotenuse and x+3 on the opposite and just x on the adjacent. I cant figure what to do with the 29 after it is given an exponent of 2 and taken out of square form.

You need to solve $\displaystyle x^2+(x+3)^2=29$.

3. ## Re: Unsure of how to solve this right triangle

yeah, but what do with 3x+6x+9=29?I am not sure how to solve this quadratic expression. I know 3x is raised to the second power, just can't figure how to use the tools on this site yet.

4. ## Re: Unsure of how to solve this right triangle

Originally Posted by goldbug78
yeah, but what do with 3x+6x+9=29?I am not sure how to solve this quadratic expression. I know 3x is squared, just can't figure how to use the tools on this site yet.
$\displaystyle \\x^2+(x+3)^2=29\\2x^2+6x+9=29\\2x^2+6x-20=0$

5. ## Re: Unsure of how to solve this right triangle

I got it now, with the answer 2. I kept thinking it was 3x instead of 2x. Thanks for the help!

7. ## Re: Unsure of how to solve this right triangle

x² + (x + 3)² = (√29)²
x² + x² + 6x + 9 = 29
2x² + 6x – 20 = 0
x² + 3x – 10 = 0
x = [– 3 ± √(3² + 40)]/2
= [– 3 ± √49]/2
= (– 3 ± 7)/2
= taking + ve values
x = (– 3 + 7)/2 = 2
sides are 2, 5, √29 units
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