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n^2+10n=-14
We move all terms to the left:
n^2+10n-(-14)=0
We add all the numbers together, and all the variables
n^2+10n+14=0
a = 1; b = 10; c = +14;
Δ = b2-4ac
Δ = 102-4·1·14
Δ = 44
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{44}=\sqrt{4*11}=\sqrt{4}*\sqrt{11}=2\sqrt{11}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{11}}{2*1}=\frac{-10-2\sqrt{11}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{11}}{2*1}=\frac{-10+2\sqrt{11}}{2} $
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