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z^2-110z+16=0
a = 1; b = -110; c = +16;
Δ = b2-4ac
Δ = -1102-4·1·16
Δ = 12036
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}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{12036}=\sqrt{4*3009}=\sqrt{4}*\sqrt{3009}=2\sqrt{3009}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-110)-2\sqrt{3009}}{2*1}=\frac{110-2\sqrt{3009}}{2} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-110)+2\sqrt{3009}}{2*1}=\frac{110+2\sqrt{3009}}{2} $
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