The area of a rectangle is given by the multiplication of its dimensions:
[tex] \text{Area} = \text{width} \times \text{length} [/tex]
In this case, we have an exact value for the area, and two variable expressions for width and length. Let's replace the formula above with what we're given:
[tex] 285 = (2x-3)(x+4) [/tex]
If you expand the right hand side, you have
[tex] 285 = 2 x^2+ 5 x-12 [/tex]
And if you move all terms to the right hand side, this becomes
[tex] 0 = 2 x^2+ 5 x-297 [/tex]
This is a quadratic equation, since it is in the form [tex]ax^2+bx+c=0 [/tex], where [tex]a = 2,\ b=5 \text{ and } c = -297 [/tex]
If you plug these values in the generic formula
[tex] x_{1,2} = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} [/tex]
you get
[tex] x = \dfrac{-27}{2} \text{ or } x = 11 [/tex]
Let's see which dimensions they yield:
[tex] x = \dfrac{-27}{2} \implies 2x-3 = -30,\quad x+4 = \dfrac{-19}{2} [/tex]
But the dimensions of a rectangle can't be negative, so we can't accept this answer.
The other solution yields
[tex] x = 11 \implies 2x-3 = 19,\quad x+4 = 15 [/tex]
So these are the dimensions of the rectangle