4(b-4)b=5

Solution for 4(b-4)b=5 equation:

4(b-4)b=5

We move all terms to the left:

4(b-4)b-(5)=0

We multiply parentheses

4b^2-16b-5=0

a = 4; b = -16; c = -5;
Δ = b2-4ac
Δ = -162-4·4·(-5)
Δ = 336
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:

$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{21}}{2*4}=\frac{16-4\sqrt{21}}{8}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{21}}{2*4}=\frac{16+4\sqrt{21}}{8}
$