-18x^2+19x-5=0

Solution for -18x^2+19x-5=0 equation:

-18x^2+19x-5=0

a = -18; b = 19; c = -5;
Δ = b2-4ac
Δ = 192-4·(-18)·(-5)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-1}{2*-18}=\frac{-20}{-36}
=5/9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+1}{2*-18}=\frac{-18}{-36}
=1/2
$