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15g+6g^2=0
a = 6; b = 15; c = 0;
Δ = b2-4ac
Δ = 152-4·6·0
Δ = 225
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{225}=15$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-15}{2*6}=\frac{-30}{12} =-2+1/2 $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+15}{2*6}=\frac{0}{12} =0 $
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