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n^2+3=6-2
We move all terms to the left:
n^2+3-(6-2)=0
We add all the numbers together, and all the variables
n^2+3-4=0
We add all the numbers together, and all the variables
n^2-1=0
a = 1; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·1·(-1)
Δ = 4
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{4}=2$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2}{2*1}=\frac{-2}{2} =-1 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2}{2*1}=\frac{2}{2} =1 $
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