Let, Length = $x$ 
Breadth = $y$
Given,
Area of Rectangle = $634{\text{m}}^2$
Length : $x$
Thrice the breadth : $3y$
2 more that thrice : $3y+2$
So, $x=3y+2$
$\text{Area of the rectangle}=\text{length}\times \text{breadth}$
$634=xy634=(2+3y)y634=2y+3y^2\therefore 3y^2+2y-634=0$
This equation resembles the general form of quadratic equation $a{x}^2+bx+c=0.$
Lets find the values of $y$ satisfying the equation.(Roots of the equation)
$y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-2\pm \sqrt{2^2-4\times 3(-634)}}{2\times 3}y=\frac{-2\pm \sqrt{4+7608}}{6}=\frac{-2\pm \sqrt{7612}}{6}y=\frac{-2\pm 87.246}{6}y=\frac{-2+87.246}{6}\text{or}\frac{-2-87.246}{6}y=14.20\text{or}-14.87$
Length is always positive.
$\therefore y=14.20\text{m}$
$x=2+3y\therefore x=2+3\left(14.20\right)=44.6\text{m}$
Length $=44.6\text{m}$ 
Breadth $=14.2\text{m}$