b^2-16b+19=0

Solution for b^2-16b+19=0 equation:

b^2-16b+19=0

a = 1; b = -16; c = +19;
Δ = b2-4ac
Δ = -162-4·1·19
Δ = 180
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-6\sqrt{5}}{2*1}=\frac{16-6\sqrt{5}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+6\sqrt{5}}{2*1}=\frac{16+6\sqrt{5}}{2}
$