n^2+21=20n

Solution for n^2+21=20n equation:

n^2+21=20n

We move all terms to the left:

n^2+21-(20n)=0

a = 1; b = -20; c = +21;
Δ = b2-4ac
Δ = -202-4·1·21
Δ = 316
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:

$\sqrt{\Delta}=\sqrt{316}=\sqrt{4*79}=\sqrt{4}*\sqrt{79}=2\sqrt{79}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{79}}{2*1}=\frac{20-2\sqrt{79}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{79}}{2*1}=\frac{20+2\sqrt{79}}{2}
$