r2-16r+45=-3

Solution for r2-16r+45=-3 equation:

r2-16r+45=-3

We move all terms to the left:

r2-16r+45-(-3)=0

We add all the numbers together, and all the variables

r^2-16r+48=0

a = 1; b = -16; c = +48;
Δ = b2-4ac
Δ = -162-4·1·48
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8}{2*1}=\frac{8}{2}
=4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8}{2*1}=\frac{24}{2}
=12
$