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20z^2-92z+96=0
a = 20; b = -92; c = +96;
Δ = b2-4ac
Δ = -922-4·20·96
Δ = 784
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{784}=28$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-92)-28}{2*20}=\frac{64}{40} =1+3/5 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-92)+28}{2*20}=\frac{120}{40} =3 $
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