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20z^2+17z-10=0
a = 20; b = 17; c = -10;
Δ = b2-4ac
Δ = 172-4·20·(-10)
Δ = 1089
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{1089}=33$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-33}{2*20}=\frac{-50}{40} =-1+1/4 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+33}{2*20}=\frac{16}{40} =2/5 $
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