20y+5y^2+4+y=0

Solution for 20y+5y^2+4+y=0 equation:

20y+5y^2+4+y=0

We add all the numbers together, and all the variables

5y^2+21y+4=0

a = 5; b = 21; c = +4;
Δ = b2-4ac
Δ = 212-4·5·4
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-19}{2*5}=\frac{-40}{10}
=-4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+19}{2*5}=\frac{-2}{10}
=-1/5
$