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19g^2+28g=0
a = 19; b = 28; c = 0;
Δ = b2-4ac
Δ = 282-4·19·0
Δ = 784
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{784}=28$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-28}{2*19}=\frac{-56}{38} =-1+9/19 $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+28}{2*19}=\frac{0}{38} =0 $
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