b^2-16b+25=0

Solution for b^2-16b+25=0 equation:

b^2-16b+25=0

a = 1; b = -16; c = +25;
Δ = b2-4ac
Δ = -162-4·1·25
Δ = 156
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:

$\sqrt{\Delta}=\sqrt{156}=\sqrt{4*39}=\sqrt{4}*\sqrt{39}=2\sqrt{39}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{39}}{2*1}=\frac{16-2\sqrt{39}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{39}}{2*1}=\frac{16+2\sqrt{39}}{2}
$