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15g^2-19g+6=0
a = 15; b = -19; c = +6;
Δ = b2-4ac
Δ = -192-4·15·6
Δ = 1
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{1}=1$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-1}{2*15}=\frac{18}{30} =3/5 $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+1}{2*15}=\frac{20}{30} =2/3 $
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