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n^2+12n=-11
We move all terms to the left:
n^2+12n-(-11)=0
We add all the numbers together, and all the variables
n^2+12n+11=0
a = 1; b = 12; c = +11;
Δ = b2-4ac
Δ = 122-4·1·11
Δ = 100
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{100}=10$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-10}{2*1}=\frac{-22}{2} =-11 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+10}{2*1}=\frac{-2}{2} =-1 $
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