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

Solution for b+3/2b+(2b-90)+(b+45)=450 equation:

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

We move all terms to the left:

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


Domain of the equation: 2b!=0

b!=0/2

b!=0

b∈R


We get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

2b^2+4b^2+2b^2-180b+90b-900b+3=0

We add all the numbers together, and all the variables

8b^2-990b+3=0

a = 8; b = -990; c = +3;
Δ = b2-4ac
Δ = -9902-4·8·3
Δ = 980004
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:

$\sqrt{\Delta}=\sqrt{980004}=\sqrt{4*245001}=\sqrt{4}*\sqrt{245001}=2\sqrt{245001}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-990)-2\sqrt{245001}}{2*8}=\frac{990-2\sqrt{245001}}{16}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-990)+2\sqrt{245001}}{2*8}=\frac{990+2\sqrt{245001}}{16}
$