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3g^2-14g+11=0
a = 3; b = -14; c = +11;
Δ = b2-4ac
Δ = -142-4·3·11
Δ = 64
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{64}=8$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-8}{2*3}=\frac{6}{6} =1 $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+8}{2*3}=\frac{22}{6} =3+2/3 $
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