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5+3/5b=7/10b
We move all terms to the left:
5+3/5b-(7/10b)=0
Domain of the equation: 5b!=0
b!=0/5
b!=0
b∈R
Domain of the equation: 10b)!=0We add all the numbers together, and all the variables
b!=0/1
b!=0
b∈R
3/5b-(+7/10b)+5=0
We get rid of parentheses
3/5b-7/10b+5=0
We calculate fractions
30b/50b^2+(-35b)/50b^2+5=0
We multiply all the terms by the denominator
30b+(-35b)+5*50b^2=0
Wy multiply elements
250b^2+30b+(-35b)=0
We get rid of parentheses
250b^2+30b-35b=0
We add all the numbers together, and all the variables
250b^2-5b=0
a = 250; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·250·0
Δ = 25
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{25}=5$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*250}=\frac{0}{500} =0 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*250}=\frac{10}{500} =1/50 $
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