l^2=96+4l

Solution for l^2=96+4l equation:

l^2=96+4l

We move all terms to the left:

l^2-(96+4l)=0

We add all the numbers together, and all the variables

l^2-(4l+96)=0

We get rid of parentheses

l^2-4l-96=0

a = 1; b = -4; c = -96;
Δ = b2-4ac
Δ = -42-4·1·(-96)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$l_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$l_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$l_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-20}{2*1}=\frac{-16}{2}
=-8
$
$l_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+20}{2*1}=\frac{24}{2}
=12
$