20x(x+1)=x-3

Solution for 20x(x+1)=x-3 equation:

20x(x+1)=x-3

We move all terms to the left:

20x(x+1)-(x-3)=0

We multiply parentheses

20x^2+20x-(x-3)=0

We get rid of parentheses

20x^2+20x-x+3=0

We add all the numbers together, and all the variables

20x^2+19x+3=0

a = 20; b = 19; c = +3;
Δ = b2-4ac
Δ = 192-4·20·3
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{-(19)-11}{2*20}=\frac{-30}{40}
=-3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+11}{2*20}=\frac{-8}{40}
=-1/5
$