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5z^2-27z-18=0
a = 5; b = -27; c = -18;
Δ = b2-4ac
Δ = -272-4·5·(-18)
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$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-33}{2*5}=\frac{-6}{10} =-3/5 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+33}{2*5}=\frac{60}{10} =6 $
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