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7z^2+27z-4=0
a = 7; b = 27; c = -4;
Δ = b2-4ac
Δ = 272-4·7·(-4)
Δ = 841
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{841}=29$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-29}{2*7}=\frac{-56}{14} =-4 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+29}{2*7}=\frac{2}{14} =1/7 $
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