.2x^2+3.2-5=0

Solution for .2x^2+3.2-5=0 equation:

.2x^2+3.2-5=0

We add all the numbers together, and all the variables

.2x^2-1.8=0

a = .2; b = 0; c = -1.8;
Δ = b2-4ac
Δ = 02-4·.2·(-1.8)
Δ = 1.44
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{-(0)-\sqrt{1.44}}{2*.2}=\frac{0-\sqrt{1.44}}{0.4}
=-\frac{\sqrt{}}{0.4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{1.44}}{2*.2}=\frac{0+\sqrt{1.44}}{0.4}
=\frac{\sqrt{}}{0.4}
$