2.01(x^2)+x=0.3

Solution for 2.01(x^2)+x=0.3 equation:

2.01(x^2)+x=0.3

We move all terms to the left:

2.01(x^2)+x-(0.3)=0

We add all the numbers together, and all the variables

2.01x^2+x-0.3=0

a = 2.01; b = 1; c = -0.3;
Δ = b2-4ac
Δ = 12-4·2.01·(-0.3)
Δ = 3.412
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{3.412}}{2*2.01}=\frac{-1-\sqrt{3.412}}{4.02}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{3.412}}{2*2.01}=\frac{-1+\sqrt{3.412}}{4.02}
$