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20+4b=1/3b+5.28

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Solution for 20+4b=1/3b+5.28 equation:



20+4b=1/3b+5.28
We move all terms to the left:
20+4b-(1/3b+5.28)=0
Domain of the equation: 3b+5.28)!=0
b∈R
We get rid of parentheses
4b-1/3b-5.28+20=0
We multiply all the terms by the denominator
4b*3b-(5.28)*3b+20*3b-1=0
We multiply parentheses
4b*3b-15.84b+20*3b-1=0
Wy multiply elements
12b^2-15.84b+60b-1=0
We add all the numbers together, and all the variables
12b^2+44.16b-1=0
a = 12; b = 44.16; c = -1;
Δ = b2-4ac
Δ = 44.162-4·12·(-1)
Δ = 1998.1056
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{-(44.16)-\sqrt{1998.1056}}{2*12}=\frac{-44.16-\sqrt{1998.1056}}{24}
b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(44.16)+\sqrt{1998.1056}}{2*12}=\frac{-44.16+\sqrt{1998.1056}}{24}

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