Question

Image attached! Please help
Really simple algebra equation!
Giving 50points!

Answer 1

Apply Pythagorean theorem

• (a-5)²+(a+2)²=(a+3)²
• a²-10a+25+a²+4a+4=a²+6a+9
• a²-6a+29=6a+9
• a²-12a+20=0
• a²-10a-2a+20=0
• a(a-10)-2(a-10)=0
• (a-2)(a-10)=0
• a=2,10

There is a side a-5 so 2 is not a answer

• a=10

Answer 2

a=10

Explanation:

use a^2+b^2=c^2

(a+2)^2+(a-5)^2=(a+3)^2

a^2+4a+4+a^2-10a+25=a^2+6a+9

2a^2-6a+29=a^2+6a+9

a^2-12a=-20

a^2-12a+20=0

quadratic formula→(12+/-8)/2

a=10 or 2

2 is not an option because 2-5=-3 which isn't possible in this case.

so a=10