1,8x^2+-2,0x+-3,2=0

Solution for 1,8x^2+-2,0x+-3,2=0 equation:

1.8x^2+-2.0x+-3.2=0

We add all the numbers together, and all the variables

1.8x^2-2x=0

a = 1.8; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·1.8·0
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{-(-2)-2}{2*1.8}=\frac{0}{3.6}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*1.8}=\frac{4}{3.6}
=1+0.4/3.6
$