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