n^2-11n+28=4

Solution for n^2-11n+28=4 equation:

n^2-11n+28=4

We move all terms to the left:

n^2-11n+28-(4)=0

We add all the numbers together, and all the variables

n^2-11n+24=0

a = 1; b = -11; c = +24;
Δ = b2-4ac
Δ = -112-4·1·24
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-5}{2*1}=\frac{6}{2}
=3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+5}{2*1}=\frac{16}{2}
=8
$