20x^2+23x+3=0

Solution for 20x^2+23x+3=0 equation:

20x^2+23x+3=0

a = 20; b = 23; c = +3;
Δ = b2-4ac
Δ = 232-4·20·3
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-17}{2*20}=\frac{-40}{40}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+17}{2*20}=\frac{-6}{40}
=-3/20
$