2/3b+b+(b+45)+(2b-90)+90=180

Solution for 2/3b+b+(b+45)+(2b-90)+90=180 equation:

2/3b+b+(b+45)+(2b-90)+90=180

We move all terms to the left:

2/3b+b+(b+45)+(2b-90)+90-(180)=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+2/3b+(b+45)+(2b-90)-90=0

We get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

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

We add all the numbers together, and all the variables

12b^2-405b+2=0

a = 12; b = -405; c = +2;
Δ = b2-4ac
Δ = -4052-4·12·2
Δ = 163929
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{-(-405)-\sqrt{163929}}{2*12}=\frac{405-\sqrt{163929}}{24}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-405)+\sqrt{163929}}{2*12}=\frac{405+\sqrt{163929}}{24}
$