b2-12b+29=-3

Solution for b2-12b+29=-3 equation:

b2-12b+29=-3

We move all terms to the left:

b2-12b+29-(-3)=0

We add all the numbers together, and all the variables

b^2-12b+32=0

a = 1; b = -12; c = +32;
Δ = b2-4ac
Δ = -122-4·1·32
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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