20x(x+2)=19x-4

Solution for 20x(x+2)=19x-4 equation:

20x(x+2)=19x-4

We move all terms to the left:

20x(x+2)-(19x-4)=0

We multiply parentheses

20x^2+40x-(19x-4)=0

We get rid of parentheses

20x^2+40x-19x+4=0

We add all the numbers together, and all the variables

20x^2+21x+4=0

a = 20; b = 21; c = +4;
Δ = b2-4ac
Δ = 212-4·20·4
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-11}{2*20}=\frac{-32}{40}
=-4/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+11}{2*20}=\frac{-10}{40}
=-1/4
$