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

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

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

We move all terms to the left:

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

We multiply parentheses

4b-(2(4b+1)b)+10=0


We calculate terms in parentheses: -(2(4b+1)b), so:

2(4b+1)b

We multiply parentheses

8b^2+2b

Back to the equation:

-(8b^2+2b)


We get rid of parentheses

-8b^2+4b-2b+10=0

We add all the numbers together, and all the variables

-8b^2+2b+10=0

a = -8; b = 2; c = +10;
Δ = b2-4ac
Δ = 22-4·(-8)·10
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-18}{2*-8}=\frac{-20}{-16}
=1+1/4
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+18}{2*-8}=\frac{16}{-16}
=-1
$