0.3x^2+2.1x+3=x^2

Solution for 0.3x^2+2.1x+3=x^2 equation:

0.3x^2+2.1x+3=x^2

We move all terms to the left:

0.3x^2+2.1x+3-(x^2)=0

determiningTheFunctionDomain
0.3x^2-x^2+2.1x+3=0

We add all the numbers together, and all the variables

-0.7x^2+2.1x+3=0

a = -0.7; b = 2.1; c = +3;
Δ = b2-4ac
Δ = 2.12-4·(-0.7)·3
Δ = 12.81
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{-(2.1)-\sqrt{12.81}}{2*-0.7}=\frac{-2.1-\sqrt{12.81}}{-1.4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2.1)+\sqrt{12.81}}{2*-0.7}=\frac{-2.1+\sqrt{12.81}}{-1.4}
$