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540=b+3/2b+90+b+45+2b-90
We move all terms to the left:
540-(b+3/2b+90+b+45+2b-90)=0
Domain of the equation: 2b+90+b+45+2b-90)!=0We add all the numbers together, and all the variables
We move all terms containing b to the left, all other terms to the right
2b+b+2b-90)!=-135
b∈R
-(4b+3/2b+45)+540=0
We get rid of parentheses
-4b-3/2b-45+540=0
We multiply all the terms by the denominator
-4b*2b-45*2b+540*2b-3=0
Wy multiply elements
-8b^2-90b+1080b-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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