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+90+(B+45)+(2B-90)+3/2B=540
We move all terms to the left:
+90+(B+45)+(2B-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+45)+(2B-90)+3/2B-450=0
We get rid of parentheses
B+2B+3/2B+45-90-450=0
We multiply all the terms by the denominator
B*2B+2B*2B+45*2B-90*2B-450*2B+3=0
Wy multiply elements
2B^2+4B^2+90B-180B-900B+3=0
We add all the numbers together, and all the variables
6B^2-990B+3=0
a = 6; b = -990; c = +3;
Δ = b2-4ac
Δ = -9902-4·6·3
Δ = 980028
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}$$B_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{980028}=\sqrt{36*27223}=\sqrt{36}*\sqrt{27223}=6\sqrt{27223}$$B_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-990)-6\sqrt{27223}}{2*6}=\frac{990-6\sqrt{27223}}{12} $$B_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-990)+6\sqrt{27223}}{2*6}=\frac{990+6\sqrt{27223}}{12} $
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