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11s^2=23s
We move all terms to the left:
11s^2-(23s)=0
a = 11; b = -23; c = 0;
Δ = b2-4ac
Δ = -232-4·11·0
Δ = 529
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{529}=23$$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-23}{2*11}=\frac{0}{22} =0 $$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+23}{2*11}=\frac{46}{22} =2+1/11 $
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