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1/b=12/3b+4
We move all terms to the left:
1/b-(12/3b+4)=0
Domain of the equation: b!=0
b∈R
Domain of the equation: 3b+4)!=0We get rid of parentheses
b∈R
1/b-12/3b-4=0
We calculate fractions
3b/3b^2+(-12b)/3b^2-4=0
We multiply all the terms by the denominator
3b+(-12b)-4*3b^2=0
Wy multiply elements
-12b^2+3b+(-12b)=0
We get rid of parentheses
-12b^2+3b-12b=0
We add all the numbers together, and all the variables
-12b^2-9b=0
a = -12; b = -9; c = 0;
Δ = b2-4ac
Δ = -92-4·(-12)·0
Δ = 81
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}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{81}=9$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-9}{2*-12}=\frac{0}{-24} =0 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+9}{2*-12}=\frac{18}{-24} =-3/4 $
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