r2-16r+63=0

Solution for r2-16r+63=0 equation:

r2-16r+63=0

We add all the numbers together, and all the variables

r^2-16r+63=0

a = 1; b = -16; c = +63;
Δ = b2-4ac
Δ = -162-4·1·63
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2}{2*1}=\frac{14}{2}
=7
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2}{2*1}=\frac{18}{2}
=9
$