n^2-14n+49=3n-21

Solution for n^2-14n+49=3n-21 equation:

n^2-14n+49=3n-21

We move all terms to the left:

n^2-14n+49-(3n-21)=0

We get rid of parentheses

n^2-14n-3n+21+49=0

We add all the numbers together, and all the variables

n^2-17n+70=0

a = 1; b = -17; c = +70;
Δ = b2-4ac
Δ = -172-4·1·70
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-3}{2*1}=\frac{14}{2}
=7
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+3}{2*1}=\frac{20}{2}
=10
$