4b(2b-3)+8b=5

Solution for 4b(2b-3)+8b=5 equation:

4b(2b-3)+8b=5

We move all terms to the left:

4b(2b-3)+8b-(5)=0

We add all the numbers together, and all the variables

8b+4b(2b-3)-5=0

We multiply parentheses

8b^2+8b-12b-5=0

We add all the numbers together, and all the variables

8b^2-4b-5=0

a = 8; b = -4; c = -5;
Δ = b2-4ac
Δ = -42-4·8·(-5)
Δ = 176
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:

$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{11}}{2*8}=\frac{4-4\sqrt{11}}{16}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{11}}{2*8}=\frac{4+4\sqrt{11}}{16}
$