4x^2+20x=9x^2

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

4x^2+20x=9x^2

We move all terms to the left:

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

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

We add all the numbers together, and all the variables

-5x^2+20x=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:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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