9b^2+20b=4b^2

Solution for 9b^2+20b=4b^2 equation:

9b^2+20b=4b^2

We move all terms to the left:

9b^2+20b-(4b^2)=0

determiningTheFunctionDomain
9b^2-4b^2+20b=0

We add all the numbers together, and all the variables

5b^2+20b=0

a = 5; b = 20; c = 0;
Δ = b2-4ac
Δ = 202-4·5·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20}{2*5}=\frac{-40}{10}
=-4
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20}{2*5}=\frac{0}{10}
=0
$