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n^2-13n+4=0
a = 1; b = -13; c = +4;
Δ = b2-4ac
Δ = -132-4·1·4
Δ = 153
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{153}=\sqrt{9*17}=\sqrt{9}*\sqrt{17}=3\sqrt{17}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-3\sqrt{17}}{2*1}=\frac{13-3\sqrt{17}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+3\sqrt{17}}{2*1}=\frac{13+3\sqrt{17}}{2} $
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