0.2x^2+2x+1.8=0

Solution for 0.2x^2+2x+1.8=0 equation:

0.2x^2+2x+1.8=0

a = 0.2; b = 2; c = +1.8;
Δ = b2-4ac
Δ = 22-4·0.2·1.8
Δ = 2.56
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{-(2)-\sqrt{2.56}}{2*0.2}=\frac{-2-\sqrt{2.56}}{0.4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+\sqrt{2.56}}{2*0.2}=\frac{-2+\sqrt{2.56}}{0.4}
$