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

Solution for 16b(b-3)+(b-3)=0 equation:

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

We multiply parentheses

16b^2-48b+(b-3)=0

We get rid of parentheses

16b^2-48b+b-3=0

We add all the numbers together, and all the variables

16b^2-47b-3=0

a = 16; b = -47; c = -3;
Δ = b2-4ac
Δ = -472-4·16·(-3)
Δ = 2401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2401}=49$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-47)-49}{2*16}=\frac{-2}{32}
=-1/16
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-47)+49}{2*16}=\frac{96}{32}
=3
$