b^2-24b+16=0

Solution for b^2-24b+16=0 equation:

b^2-24b+16=0

a = 1; b = -24; c = +16;
Δ = b2-4ac
Δ = -242-4·1·16
Δ = 512
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:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-16\sqrt{2}}{2*1}=\frac{24-16\sqrt{2}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+16\sqrt{2}}{2*1}=\frac{24+16\sqrt{2}}{2}
$