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20z^2+11z-4=0
a = 20; b = 11; c = -4;
Δ = b2-4ac
Δ = 112-4·20·(-4)
Δ = 441
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{441}=21$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-21}{2*20}=\frac{-32}{40} =-4/5 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+21}{2*20}=\frac{10}{40} =1/4 $
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