.2x^2+x=1.2

Solution for .2x^2+x=1.2 equation:

.2x^2+x=1.2

We move all terms to the left:

.2x^2+x-(1.2)=0

We add all the numbers together, and all the variables

.2x^2+x-1.2=0

a = .2; b = 1; c = -1.2;
Δ = b2-4ac
Δ = 12-4·.2·(-1.2)
Δ = 1.96
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{-(1)-\sqrt{1.96}}{2*.2}=\frac{-1-\sqrt{1.96}}{0.4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{1.96}}{2*.2}=\frac{-1+\sqrt{1.96}}{0.4}
$