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n^2+6n-45=0
a = 1; b = 6; c = -45;
Δ = b2-4ac
Δ = 62-4·1·(-45)
Δ = 216
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{6}}{2*1}=\frac{-6-6\sqrt{6}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{6}}{2*1}=\frac{-6+6\sqrt{6}}{2} $
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