(2b-90)+90+(b+45)+2/3b+b=540

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

(2b-90)+90+(b+45)+2/3b+b=540

We move all terms to the left:

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


Domain of the equation: 3b!=0

b!=0/3

b!=0

b∈R


We add all the numbers together, and all the variables

b+(2b-90)+(b+45)+2/3b-450=0

We get rid of parentheses

b+2b+b+2/3b-90+45-450=0

We multiply all the terms by the denominator


b*3b+2b*3b+b*3b-90*3b+45*3b-450*3b+2=0

Wy multiply elements

3b^2+6b^2+3b^2-270b+135b-1350b+2=0

We add all the numbers together, and all the variables

12b^2-1485b+2=0

a = 12; b = -1485; c = +2;
Δ = b2-4ac
Δ = -14852-4·12·2
Δ = 2205129
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1485)-\sqrt{2205129}}{2*12}=\frac{1485-\sqrt{2205129}}{24}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1485)+\sqrt{2205129}}{2*12}=\frac{1485+\sqrt{2205129}}{24}
$