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

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

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

We move all terms to the left:

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

We multiply parentheses

-3b+5b-(4b(3b+5))+10=0


We calculate terms in parentheses: -(4b(3b+5)), so:

4b(3b+5)

We multiply parentheses

12b^2+20b

Back to the equation:

-(12b^2+20b)


We add all the numbers together, and all the variables

2b-(12b^2+20b)+10=0

We get rid of parentheses

-12b^2+2b-20b+10=0

We add all the numbers together, and all the variables

-12b^2-18b+10=0

a = -12; b = -18; c = +10;
Δ = b2-4ac
Δ = -182-4·(-12)·10
Δ = 804
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:

$\sqrt{\Delta}=\sqrt{804}=\sqrt{4*201}=\sqrt{4}*\sqrt{201}=2\sqrt{201}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{201}}{2*-12}=\frac{18-2\sqrt{201}}{-24}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{201}}{2*-12}=\frac{18+2\sqrt{201}}{-24}
$