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28g^2-39g=0
a = 28; b = -39; c = 0;
Δ = b2-4ac
Δ = -392-4·28·0
Δ = 1521
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{1521}=39$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-39)-39}{2*28}=\frac{0}{56} =0 $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+39}{2*28}=\frac{78}{56} =1+11/28 $
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