b^2-9b+19=5

Solution for b^2-9b+19=5 equation:

b^2-9b+19=5

We move all terms to the left:

b^2-9b+19-(5)=0

We add all the numbers together, and all the variables

b^2-9b+14=0

a = 1; b = -9; c = +14;
Δ = b2-4ac
Δ = -92-4·1·14
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-5}{2*1}=\frac{4}{2}
=2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+5}{2*1}=\frac{14}{2}
=7
$