b^2-18b+16=0

Solution for b^2-18b+16=0 equation:

b^2-18b+16=0

a = 1; b = -18; c = +16;
Δ = b2-4ac
Δ = -182-4·1·16
Δ = 260
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:

$\sqrt{\Delta}=\sqrt{260}=\sqrt{4*65}=\sqrt{4}*\sqrt{65}=2\sqrt{65}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{65}}{2*1}=\frac{18-2\sqrt{65}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{65}}{2*1}=\frac{18+2\sqrt{65}}{2}
$