(20w^2+21w+4)=0

Solution for (20w^2+21w+4)=0 equation:

(20w^2+21w+4)=0

We get rid of parentheses

20w^2+21w+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:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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