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g^2-19g+50=0
a = 1; b = -19; c = +50;
Δ = b2-4ac
Δ = -192-4·1·50
Δ = 161
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-\sqrt{161}}{2*1}=\frac{19-\sqrt{161}}{2} $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{161}}{2*1}=\frac{19+\sqrt{161}}{2} $
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