20x^2+5x=4x

Solution for 20x^2+5x=4x equation:

20x^2+5x=4x

We move all terms to the left:

20x^2+5x-(4x)=0

We add all the numbers together, and all the variables

20x^2+x=0

a = 20; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·20·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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