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g^2=1024
We move all terms to the left:
g^2-(1024)=0
a = 1; b = 0; c = -1024;
Δ = b2-4ac
Δ = 02-4·1·(-1024)
Δ = 4096
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{4096}=64$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-64}{2*1}=\frac{-64}{2} =-32 $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+64}{2*1}=\frac{64}{2} =32 $
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