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4z^2+42z+20=0
a = 4; b = 42; c = +20;
Δ = b2-4ac
Δ = 422-4·4·20
Δ = 1444
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{1444}=38$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-38}{2*4}=\frac{-80}{8} =-10 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+38}{2*4}=\frac{-4}{8} =-1/2 $
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