20x^2+19x+3=0.

Solution for 20x^2+19x+3=0. equation:

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
$