b^2-16b+45=0

Solution for b^2-16b+45=0 equation:

b^2-16b+45=0

a = 1; b = -16; c = +45;
Δ = b2-4ac
Δ = -162-4·1·45
Δ = 76
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:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{19}}{2*1}=\frac{16-2\sqrt{19}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{19}}{2*1}=\frac{16+2\sqrt{19}}{2}
$