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3g^2+47g-16=0
a = 3; b = 47; c = -16;
Δ = b2-4ac
Δ = 472-4·3·(-16)
Δ = 2401
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{2401}=49$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(47)-49}{2*3}=\frac{-96}{6} =-16 $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(47)+49}{2*3}=\frac{2}{6} =1/3 $
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