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

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

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

We move all terms to the left:

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


Domain of the equation: 2b!=0

b!=0/2

b!=0

b∈R


We get rid of parentheses

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

We multiply all the terms by the denominator


b*2b+b*2b+2b*2b+45*2b+90*2b-540*2b+3=0

Wy multiply elements

2b^2+2b^2+4b^2+90b+180b-1080b+3=0

We add all the numbers together, and all the variables

8b^2-810b+3=0

a = 8; b = -810; c = +3;
Δ = b2-4ac
Δ = -8102-4·8·3
Δ = 656004
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:

$\sqrt{\Delta}=\sqrt{656004}=\sqrt{4*164001}=\sqrt{4}*\sqrt{164001}=2\sqrt{164001}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-810)-2\sqrt{164001}}{2*8}=\frac{810-2\sqrt{164001}}{16}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-810)+2\sqrt{164001}}{2*8}=\frac{810+2\sqrt{164001}}{16}
$