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3g^2+26g+16=0
a = 3; b = 26; c = +16;
Δ = b2-4ac
Δ = 262-4·3·16
Δ = 484
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{484}=22$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-22}{2*3}=\frac{-48}{6} =-8 $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+22}{2*3}=\frac{-4}{6} =-2/3 $
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