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(s^2)+19s-12=0
a = 1; b = 19; c = -12;
Δ = b2-4ac
Δ = 192-4·1·(-12)
Δ = 409
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}$$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-\sqrt{409}}{2*1}=\frac{-19-\sqrt{409}}{2} $$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{409}}{2*1}=\frac{-19+\sqrt{409}}{2} $
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