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b=3/5b
We move all terms to the left:
b-(3/5b)=0
Domain of the equation: 5b)!=0We add all the numbers together, and all the variables
b!=0/1
b!=0
b∈R
b-(+3/5b)=0
We get rid of parentheses
b-3/5b=0
We multiply all the terms by the denominator
b*5b-3=0
Wy multiply elements
5b^2-3=0
a = 5; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·5·(-3)
Δ = 60
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}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{15}}{2*5}=\frac{0-2\sqrt{15}}{10} =-\frac{2\sqrt{15}}{10} =-\frac{\sqrt{15}}{5} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{15}}{2*5}=\frac{0+2\sqrt{15}}{10} =\frac{2\sqrt{15}}{10} =\frac{\sqrt{15}}{5} $
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