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(n-7)(n=3)
We move all terms to the left:
(n-7)(n-(3))=0
We multiply parentheses ..
(+n^2-3n-7n+21)=0
We get rid of parentheses
n^2-3n-7n+21=0
We add all the numbers together, and all the variables
n^2-10n+21=0
a = 1; b = -10; c = +21;
Δ = b2-4ac
Δ = -102-4·1·21
Δ = 16
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{16}=4$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-4}{2*1}=\frac{6}{2} =3 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+4}{2*1}=\frac{14}{2} =7 $
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