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-n^2+12=n
We move all terms to the left:
-n^2+12-(n)=0
We add all the numbers together, and all the variables
-1n^2-1n+12=0
a = -1; b = -1; c = +12;
Δ = b2-4ac
Δ = -12-4·(-1)·12
Δ = 49
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{49}=7$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-7}{2*-1}=\frac{-6}{-2} =+3 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+7}{2*-1}=\frac{8}{-2} =-4 $
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