3+x=x^2+3x

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

3+x=x^2+3x

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1x^2-2x+3=0

a = -1; b = -2; c = +3;
Δ = b2-4ac
Δ = -22-4·(-1)·3
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-4}{2*-1}=\frac{-2}{-2}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+4}{2*-1}=\frac{6}{-2}
=-3
$