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2g^2-27g+13=0
a = 2; b = -27; c = +13;
Δ = b2-4ac
Δ = -272-4·2·13
Δ = 625
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{625}=25$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-25}{2*2}=\frac{2}{4} =1/2 $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+25}{2*2}=\frac{52}{4} =13 $
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