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23s^2+19s=0
a = 23; b = 19; c = 0;
Δ = b2-4ac
Δ = 192-4·23·0
Δ = 361
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{361}=19$$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-19}{2*23}=\frac{-38}{46} =-19/23 $$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+19}{2*23}=\frac{0}{46} =0 $
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