0=l(l+4)-48

Solution for 0=l(l+4)-48 equation:

0=l(l+4)-48

We move all terms to the left:

0-(l(l+4)-48)=0

We add all the numbers together, and all the variables

-(l(l+4)-48)=0


We calculate terms in parentheses: -(l(l+4)-48), so:

l(l+4)-48

We multiply parentheses

l^2+4l-48

Back to the equation:

-(l^2+4l-48)


We get rid of parentheses

-l^2-4l+48=0

We add all the numbers together, and all the variables

-1l^2-4l+48=0

a = -1; b = -4; c = +48;
Δ = b2-4ac
Δ = -42-4·(-1)·48
Δ = 208
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$l_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$l_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$l_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{13}}{2*-1}=\frac{4-4\sqrt{13}}{-2}
$
$l_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{13}}{2*-1}=\frac{4+4\sqrt{13}}{-2}
$