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2z^2-200=0
a = 2; b = 0; c = -200;
Δ = b2-4ac
Δ = 02-4·2·(-200)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1600}=40$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40}{2*2}=\frac{-40}{4} =-10 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40}{2*2}=\frac{40}{4} =10 $
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