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