Quadratic is a type of problem which deals with a variable multiplied by itself- an operation also known as squaring.
Solution:
Let the length of one side of the right triangle be $x$ cm
So, the other side will be $=\left( x+5 \right)$ cm [as they differ by $5$cm]
And given that hypotenuse $=25$ cm
On applying Pythagoras Theorem, we have
${{x}^{2}}+{{\left( x+5 \right)}^{2}}={{25}^{2}}$
${{x}^{2}}+{{x}^{2}}+10x+25=625$
$2{{x}^{2}}+10x+25-625=0$
$2{{x}^{2}}+10x-600=0$
${{x}^{2}}+5x-300=0$
${{x}^{2}}-15x+20x-300=0$ [By factorisation method]
$x\left( x-15 \right)+20\left( x-15 \right)=0$
$\left( x-15 \right)\left( x+20 \right)=0$
$x=15$ or $x=-20$ (neglected) As the side of triangle can never be negative.
Thus, when $x=15\Rightarrow x+5=15+5=20$
Hence, the length of side of right triangle is $15$ cm and other side is $20$ cm