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b+(b+45)+(2b-90)+90+3/2b=540
We move all terms to the left:
b+(b+45)+(2b-90)+90+3/2b-(540)=0
Domain of the equation: 2b!=0We add all the numbers together, and all the variables
b!=0/2
b!=0
b∈R
b+(b+45)+(2b-90)+3/2b-450=0
We get rid of parentheses
b+b+2b+3/2b+45-90-450=0
We multiply all the terms by the denominator
b*2b+b*2b+2b*2b+45*2b-90*2b-450*2b+3=0
Wy multiply elements
2b^2+2b^2+4b^2+90b-180b-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} $
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