n^2+98=21n

Solution for n^2+98=21n equation:

n^2+98=21n

We move all terms to the left:

n^2+98-(21n)=0

a = 1; b = -21; c = +98;
Δ = b2-4ac
Δ = -212-4·1·98
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-7}{2*1}=\frac{14}{2}
=7
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+7}{2*1}=\frac{28}{2}
=14
$