B+b+45+2b-90+90+3/2b=540

Solution for B+b+45+2b-90+90+3/2b=540 equation:

+B+45+2B-90+90+3/2B=540

We move all terms to the left:

+B+45+2B-90+90+3/2B-(540)=0


Domain of the equation: 2B!=0

B!=0/2

B!=0

B∈R


We add all the numbers together, and all the variables

3B+3/2B-495=0

We multiply all the terms by the denominator


3B*2B-495*2B+3=0

Wy multiply elements

6B^2-990B+3=0

a = 6; b = -990; c = +3;
Δ = b2-4ac
Δ = -9902-4·6·3
Δ = 980028
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:

$\sqrt{\Delta}=\sqrt{980028}=\sqrt{36*27223}=\sqrt{36}*\sqrt{27223}=6\sqrt{27223}$
$B_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-990)-6\sqrt{27223}}{2*6}=\frac{990-6\sqrt{27223}}{12}
$
$B_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-990)+6\sqrt{27223}}{2*6}=\frac{990+6\sqrt{27223}}{12}
$