-4b(b+3)-12=2b-16+5b

Solution for -4b(b+3)-12=2b-16+5b equation:

-4b(b+3)-12=2b-16+5b

We move all terms to the left:

-4b(b+3)-12-(2b-16+5b)=0

We add all the numbers together, and all the variables

-4b(b+3)-(7b-16)-12=0

We multiply parentheses

-4b^2-12b-(7b-16)-12=0

We get rid of parentheses

-4b^2-12b-7b+16-12=0

We add all the numbers together, and all the variables

-4b^2-19b+4=0

a = -4; b = -19; c = +4;
Δ = b2-4ac
Δ = -192-4·(-4)·4
Δ = 425
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:

$\sqrt{\Delta}=\sqrt{425}=\sqrt{25*17}=\sqrt{25}*\sqrt{17}=5\sqrt{17}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-5\sqrt{17}}{2*-4}=\frac{19-5\sqrt{17}}{-8}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+5\sqrt{17}}{2*-4}=\frac{19+5\sqrt{17}}{-8}
$