0=20x^2+21x+4

Solution for 0=20x^2+21x+4 equation:

0=20x^2+21x+4

We move all terms to the left:

0-(20x^2+21x+4)=0

We add all the numbers together, and all the variables

-(20x^2+21x+4)=0

We get rid of parentheses

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