(n-7)(n=3)

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

(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
$