0.1x2+0.2x+0.3=3000

Solution for 0.1x2+0.2x+0.3=3000 equation:

0.1x^2+0.2x+0.3=3000

We move all terms to the left:

0.1x^2+0.2x+0.3-(3000)=0

We add all the numbers together, and all the variables

0.1x^2+0.2x-2999.7=0

a = 0.1; b = 0.2; c = -2999.7;
Δ = b2-4ac
Δ = 0.22-4·0.1·(-2999.7)
Δ = 1199.92
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{-(0.2)-\sqrt{1199.92}}{2*0.1}=\frac{-0.2-\sqrt{1199.92}}{0.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.2)+\sqrt{1199.92}}{2*0.1}=\frac{-0.2+\sqrt{1199.92}}{0.2}
$