19x^2+19x+3.55=0

Solution for 19x^2+19x+3.55=0 equation:

19x^2+19x+3.55=0

a = 19; b = 19; c = +3.55;
Δ = b2-4ac
Δ = 192-4·19·3.55
Δ = 91.2
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-\sqrt{91.2}}{2*19}=\frac{-19-\sqrt{91.2}}{38}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{91.2}}{2*19}=\frac{-19+\sqrt{91.2}}{38}
$