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9g^2+16g-4=0
a = 9; b = 16; c = -4;
Δ = b2-4ac
Δ = 162-4·9·(-4)
Δ = 400
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{400}=20$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-20}{2*9}=\frac{-36}{18} =-2 $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+20}{2*9}=\frac{4}{18} =2/9 $
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