3b(2b+3)=9b(b+9)

Solution for 3b(2b+3)=9b(b+9) equation:

3b(2b+3)=9b(b+9)

We move all terms to the left:

3b(2b+3)-(9b(b+9))=0

We multiply parentheses

6b^2+9b-(9b(b+9))=0


We calculate terms in parentheses: -(9b(b+9)), so:

9b(b+9)

We multiply parentheses

9b^2+81b

Back to the equation:

-(9b^2+81b)


We get rid of parentheses

6b^2-9b^2+9b-81b=0

We add all the numbers together, and all the variables

-3b^2-72b=0

a = -3; b = -72; c = 0;
Δ = b2-4ac
Δ = -722-4·(-3)·0
Δ = 5184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5184}=72$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-72)-72}{2*-3}=\frac{0}{-6}
=0
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-72)+72}{2*-3}=\frac{144}{-6}
=-24
$