4b(2b+5)=2(3b-1)

Solution for 4b(2b+5)=2(3b-1) equation:

4b(2b+5)=2(3b-1)

We move all terms to the left:

4b(2b+5)-(2(3b-1))=0

We multiply parentheses

8b^2+20b-(2(3b-1))=0


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

2(3b-1)

We multiply parentheses

6b-2

Back to the equation:

-(6b-2)


We get rid of parentheses

8b^2+20b-6b+2=0

We add all the numbers together, and all the variables

8b^2+14b+2=0

a = 8; b = 14; c = +2;
Δ = b2-4ac
Δ = 142-4·8·2
Δ = 132
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:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{33}}{2*8}=\frac{-14-2\sqrt{33}}{16}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{33}}{2*8}=\frac{-14+2\sqrt{33}}{16}
$