18x^2+19x+1=0

Solution for 18x^2+19x+1=0 equation:

18x^2+19x+1=0

a = 18; b = 19; c = +1;
Δ = b2-4ac
Δ = 192-4·18·1
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{-(19)-17}{2*18}=\frac{-36}{36}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+17}{2*18}=\frac{-2}{36}
=-1/18
$