3/2b+(b+45)+(2b-40)+90+b=540

Solution for 3/2b+(b+45)+(2b-40)+90+b=540 equation:

3/2b+(b+45)+(2b-40)+90+b=540

We move all terms to the left:

3/2b+(b+45)+(2b-40)+90+b-(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

b+3/2b+(b+45)+(2b-40)-450=0

We get rid of parentheses

b+3/2b+b+2b+45-40-450=0

We multiply all the terms by the denominator


b*2b+b*2b+2b*2b+45*2b-40*2b-450*2b+3=0

Wy multiply elements

2b^2+2b^2+4b^2+90b-80b-900b+3=0

We add all the numbers together, and all the variables

8b^2-890b+3=0

a = 8; b = -890; c = +3;
Δ = b2-4ac
Δ = -8902-4·8·3
Δ = 792004
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:

$\sqrt{\Delta}=\sqrt{792004}=\sqrt{4*198001}=\sqrt{4}*\sqrt{198001}=2\sqrt{198001}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-890)-2\sqrt{198001}}{2*8}=\frac{890-2\sqrt{198001}}{16}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-890)+2\sqrt{198001}}{2*8}=\frac{890+2\sqrt{198001}}{16}
$