n^2-4n+7=21

Solution for n^2-4n+7=21 equation:

n^2-4n+7=21

We move all terms to the left:

n^2-4n+7-(21)=0

We add all the numbers together, and all the variables

n^2-4n-14=0

a = 1; b = -4; c = -14;
Δ = b2-4ac
Δ = -42-4·1·(-14)
Δ = 72
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-6\sqrt{2}}{2*1}=\frac{4-6\sqrt{2}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+6\sqrt{2}}{2*1}=\frac{4+6\sqrt{2}}{2}
$