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g^2-12g+11=0
a = 1; b = -12; c = +11;
Δ = b2-4ac
Δ = -122-4·1·11
Δ = 100
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{100}=10$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-10}{2*1}=\frac{2}{2} =1 $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+10}{2*1}=\frac{22}{2} =11 $
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