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g^2=100

We move all terms to the left:

g^2-(100)=0

a = 1; b = 0; c = -100;

Δ = b^{2}-4ac

Δ = 0^{2}-4·1·(-100)

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$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20}{2*1}=\frac{-20}{2} =-10 $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20}{2*1}=\frac{20}{2} =10 $

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