Please help? Question on the Pythagoras theorem

the lengths of the sides of a triangle are xcm, (x+1)cm and (x+2) cm. Determine x so that this triangle is a right angled triangle.

And please explain

Please help :)
Thanks :)

solution;
for right angled triangle
hypotenuse squared=perpendicular squared + base squared
here x+2 will be hypotenuse because x is some length it cant be negative and x+2 will be the greatest side
then by pythagorus theorm
we have
[x+2]^2=x^2 + [x+1]^2
or, [x+2]^2-[x+1]^2=x^2
or, [x+2+x+1][x+2-x-1]=x^2
or, 2x+3=x^2
or, x^2-2x-3=0
or,x^2-3x+x-3=0
or,x[x-3]+1[x-3]=0
or,[x-3][x+1]=0
that implies
either x=3 or x = -1
-1 cant be value of x because length is not negative
thats why x=3,x+1=4,x+2=5
.these are the values..

Assuming x to be a +ve quantity.Then x+2 will be the hypotnuse.

x^2 +(x+1)^2=(x+2)^2
=> x^2 +(x^2+2x+1)=(x^2 +4x+4)
=> x^2-2x-3=0
=> (x-3)(x+1)=0
=> x=3;

X=1cm... apply pythagorus, (x+2)^2 = (x+1)^2+x^2


Solve this, you'll get 1 and -3, as distance cannot be negative so 1...

x^2 + ( x + 1) ^2 = (x + 2)^ 2

x^2 -2x -3 = 0

(x-3)( x +1) = 0

x = 3

x = -1

The length must be positive

Therefore, x = 3