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s^2=3s-2
We move all terms to the left:
s^2-(3s-2)=0
We get rid of parentheses
s^2-3s+2=0
a = 1; b = -3; c = +2;
Δ = b2-4ac
Δ = -32-4·1·2
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1}=1$$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-1}{2*1}=\frac{2}{2} =1 $$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+1}{2*1}=\frac{4}{2} =2 $
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