n^2+68=21n

Solution for n^2+68=21n equation:

n^2+68=21n

We move all terms to the left:

n^2+68-(21n)=0

a = 1; b = -21; c = +68;
Δ = b2-4ac
Δ = -212-4·1·68
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-13}{2*1}=\frac{8}{2}
=4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+13}{2*1}=\frac{34}{2}
=17
$