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z^2-18z+56=0
a = 1; b = -18; c = +56;
Δ = b2-4ac
Δ = -182-4·1·56
Δ = 100
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{100}=10$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-10}{2*1}=\frac{8}{2} =4 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+10}{2*1}=\frac{28}{2} =14 $
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