(n+3)(n-7)=

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Solution for (n+3)(n-7)= equation:



(n+3)(n-7)=
We move all terms to the left:
(n+3)(n-7)-()=0
We add all the numbers together, and all the variables
(n+3)(n-7)=0
We multiply parentheses ..
(+n^2-7n+3n-21)=0
We get rid of parentheses
n^2-7n+3n-21=0
We add all the numbers together, and all the variables
n^2-4n-21=0
a = 1; b = -4; c = -21;
Δ = b2-4ac
Δ = -42-4·1·(-21)
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{-(-4)-10}{2*1}=\frac{-6}{2} =-3 $
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+10}{2*1}=\frac{14}{2} =7 $

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