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n^2-14n-60=0
a = 1; b = -14; c = -60;
Δ = b2-4ac
Δ = -142-4·1·(-60)
Δ = 436
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{436}=\sqrt{4*109}=\sqrt{4}*\sqrt{109}=2\sqrt{109}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{109}}{2*1}=\frac{14-2\sqrt{109}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{109}}{2*1}=\frac{14+2\sqrt{109}}{2} $
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