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(b-3)(b-1)=0
We multiply parentheses ..
(+b^2-1b-3b+3)=0
We get rid of parentheses
b^2-1b-3b+3=0
We add all the numbers together, and all the variables
b^2-4b+3=0
a = 1; b = -4; c = +3;
Δ = b2-4ac
Δ = -42-4·1·3
Δ = 4
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{4}=2$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2}{2*1}=\frac{2}{2} =1 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2}{2*1}=\frac{6}{2} =3 $
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