-4b(5-3b)=3b+16

Solution for -4b(5-3b)=3b+16 equation:

-4b(5-3b)=3b+16

We move all terms to the left:

-4b(5-3b)-(3b+16)=0

We add all the numbers together, and all the variables

-4b(-3b+5)-(3b+16)=0

We multiply parentheses

12b^2-20b-(3b+16)=0

We get rid of parentheses

12b^2-20b-3b-16=0

We add all the numbers together, and all the variables

12b^2-23b-16=0

a = 12; b = -23; c = -16;
Δ = b2-4ac
Δ = -232-4·12·(-16)
Δ = 1297
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-\sqrt{1297}}{2*12}=\frac{23-\sqrt{1297}}{24}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{1297}}{2*12}=\frac{23+\sqrt{1297}}{24}
$