-4(b+1)2b=3(2b-1)

Solution for -4(b+1)2b=3(2b-1) equation:

-4(b+1)2b=3(2b-1)

We move all terms to the left:

-4(b+1)2b-(3(2b-1))=0

We multiply parentheses

-8b^2-8b-(3(2b-1))=0


We calculate terms in parentheses: -(3(2b-1)), so:

3(2b-1)

We multiply parentheses

6b-3

Back to the equation:

-(6b-3)


We get rid of parentheses

-8b^2-8b-6b+3=0

We add all the numbers together, and all the variables

-8b^2-14b+3=0

a = -8; b = -14; c = +3;
Δ = b2-4ac
Δ = -142-4·(-8)·3
Δ = 292
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:

$\sqrt{\Delta}=\sqrt{292}=\sqrt{4*73}=\sqrt{4}*\sqrt{73}=2\sqrt{73}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{73}}{2*-8}=\frac{14-2\sqrt{73}}{-16}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{73}}{2*-8}=\frac{14+2\sqrt{73}}{-16}
$