2b-5+(b-6)=2b(3-8b)

Solution for 2b-5+(b-6)=2b(3-8b) equation:

2b-5+(b-6)=2b(3-8b)

We move all terms to the left:

2b-5+(b-6)-(2b(3-8b))=0

We add all the numbers together, and all the variables

2b+(b-6)-(2b(-8b+3))-5=0

We get rid of parentheses

2b+b-(2b(-8b+3))-6-5=0


We calculate terms in parentheses: -(2b(-8b+3)), so:

2b(-8b+3)

We multiply parentheses

-16b^2+6b

Back to the equation:

-(-16b^2+6b)


We add all the numbers together, and all the variables

-(-16b^2+6b)+3b-11=0

We get rid of parentheses

16b^2-6b+3b-11=0

We add all the numbers together, and all the variables

16b^2-3b-11=0

a = 16; b = -3; c = -11;
Δ = b2-4ac
Δ = -32-4·16·(-11)
Δ = 713
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{-(-3)-\sqrt{713}}{2*16}=\frac{3-\sqrt{713}}{32}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{713}}{2*16}=\frac{3+\sqrt{713}}{32}
$