20x^2+21x+1=0

Solution for 20x^2+21x+1=0 equation:

20x^2+21x+1=0

a = 20; b = 21; c = +1;
Δ = b2-4ac
Δ = 212-4·20·1
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-19}{2*20}=\frac{-40}{40}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+19}{2*20}=\frac{-2}{40}
=-1/20
$