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400x^2=1
We move all terms to the left:
400x^2-(1)=0
a = 400; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·400·(-1)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1600}=40$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40}{2*400}=\frac{-40}{800} =-1/20 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40}{2*400}=\frac{40}{800} =1/20 $
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