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

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

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

We move all terms to the left:

+3/2B+B+45+2B+90+90-(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-315=0

We multiply all the terms by the denominator


3B*2B-315*2B+3=0

Wy multiply elements

6B^2-630B+3=0

a = 6; b = -630; c = +3;
Δ = b2-4ac
Δ = -6302-4·6·3
Δ = 396828
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:

$\sqrt{\Delta}=\sqrt{396828}=\sqrt{36*11023}=\sqrt{36}*\sqrt{11023}=6\sqrt{11023}$
$B_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-630)-6\sqrt{11023}}{2*6}=\frac{630-6\sqrt{11023}}{12}
$
$B_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-630)+6\sqrt{11023}}{2*6}=\frac{630+6\sqrt{11023}}{12}
$