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9g^2+18g-16=0
a = 9; b = 18; c = -16;
Δ = b2-4ac
Δ = 182-4·9·(-16)
Δ = 900
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{900}=30$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-30}{2*9}=\frac{-48}{18} =-2+2/3 $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+30}{2*9}=\frac{12}{18} =2/3 $
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