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5/3b+5=20-b
We move all terms to the left:
5/3b+5-(20-b)=0
Domain of the equation: 3b!=0We add all the numbers together, and all the variables
b!=0/3
b!=0
b∈R
5/3b-(-1b+20)+5=0
We get rid of parentheses
5/3b+1b-20+5=0
We multiply all the terms by the denominator
1b*3b-20*3b+5*3b+5=0
Wy multiply elements
3b^2-60b+15b+5=0
We add all the numbers together, and all the variables
3b^2-45b+5=0
a = 3; b = -45; c = +5;
Δ = b2-4ac
Δ = -452-4·3·5
Δ = 1965
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{-(-45)-\sqrt{1965}}{2*3}=\frac{45-\sqrt{1965}}{6} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+\sqrt{1965}}{2*3}=\frac{45+\sqrt{1965}}{6} $
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