2x^2+1.6x-0.18=0

Solution for 2x^2+1.6x-0.18=0 equation:

2x^2+1.6x-0.18=0

a = 2; b = 1.6; c = -0.18;
Δ = b2-4ac
Δ = 1.62-4·2·(-0.18)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1.6)-2}{2*2}=\frac{-3.6}{4}
=-2.25/2.5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.6)+2}{2*2}=\frac{0.4}{4}
=0.4/4
$