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n^2-3n-144=0
a = 1; b = -3; c = -144;
Δ = b2-4ac
Δ = -32-4·1·(-144)
Δ = 585
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{585}=\sqrt{9*65}=\sqrt{9}*\sqrt{65}=3\sqrt{65}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{65}}{2*1}=\frac{3-3\sqrt{65}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{65}}{2*1}=\frac{3+3\sqrt{65}}{2} $
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