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z^2-25z+144=0
a = 1; b = -25; c = +144;
Δ = b2-4ac
Δ = -252-4·1·144
Δ = 49
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{49}=7$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-7}{2*1}=\frac{18}{2} =9 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+7}{2*1}=\frac{32}{2} =16 $
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