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-1/2b+3b=0
Domain of the equation: 2b!=0We add all the numbers together, and all the variables
b!=0/2
b!=0
b∈R
3b-1/2b=0
We multiply all the terms by the denominator
3b*2b-1=0
Wy multiply elements
6b^2-1=0
a = 6; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·6·(-1)
Δ = 24
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}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{6}}{2*6}=\frac{0-2\sqrt{6}}{12} =-\frac{2\sqrt{6}}{12} =-\frac{\sqrt{6}}{6} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{6}}{2*6}=\frac{0+2\sqrt{6}}{12} =\frac{2\sqrt{6}}{12} =\frac{\sqrt{6}}{6} $
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