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g^2-23g=0
a = 1; b = -23; c = 0;
Δ = b2-4ac
Δ = -232-4·1·0
Δ = 529
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}$$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{529}=23$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-23}{2*1}=\frac{0}{2} =0 $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+23}{2*1}=\frac{46}{2} =23 $
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