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28z^2+29z+1=0
a = 28; b = 29; c = +1;
Δ = b2-4ac
Δ = 292-4·28·1
Δ = 729
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{729}=27$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-27}{2*28}=\frac{-56}{56} =-1 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+27}{2*28}=\frac{-2}{56} =-1/28 $
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