b^2-14b+29=5

Solution for b^2-14b+29=5 equation:

b^2-14b+29=5

We move all terms to the left:

b^2-14b+29-(5)=0

We add all the numbers together, and all the variables

b^2-14b+24=0

a = 1; b = -14; c = +24;
Δ = b2-4ac
Δ = -142-4·1·24
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-10}{2*1}=\frac{4}{2}
=2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+10}{2*1}=\frac{24}{2}
=12
$