(54-1b)(48+1b)=180

Solution for (54-1b)(48+1b)=180 equation:

(54-1b)(48+1b)=180

We move all terms to the left:

(54-1b)(48+1b)-(180)=0

We add all the numbers together, and all the variables

(-1b+54)(b+48)-180=0

We multiply parentheses ..

(-1b^2-48b+54b+2592)-180=0

We get rid of parentheses

-1b^2-48b+54b+2592-180=0

We add all the numbers together, and all the variables

-1b^2+6b+2412=0

a = -1; b = 6; c = +2412;
Δ = b2-4ac
Δ = 62-4·(-1)·2412
Δ = 9684
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{9684}=\sqrt{36*269}=\sqrt{36}*\sqrt{269}=6\sqrt{269}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{269}}{2*-1}=\frac{-6-6\sqrt{269}}{-2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{269}}{2*-1}=\frac{-6+6\sqrt{269}}{-2}
$