0.6x^2+0.1x=1.2

Solution for 0.6x^2+0.1x=1.2 equation:

0.6x^2+0.1x=1.2

We move all terms to the left:

0.6x^2+0.1x-(1.2)=0

We add all the numbers together, and all the variables

0.6x^2+0.1x-1.2=0

a = 0.6; b = 0.1; c = -1.2;
Δ = b2-4ac
Δ = 0.12-4·0.6·(-1.2)
Δ = 2.89
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{-(0.1)-\sqrt{2.89}}{2*0.6}=\frac{-0.1-\sqrt{2.89}}{1.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.1)+\sqrt{2.89}}{2*0.6}=\frac{-0.1+\sqrt{2.89}}{1.2}
$