1+x^2+x=3

Solution for 1+x^2+x=3 equation:

1+x^2+x=3

We move all terms to the left:

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

We add all the numbers together, and all the variables

x^2+x-2=0

a = 1; b = 1; c = -2;
Δ = b2-4ac
Δ = 12-4·1·(-2)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3}{2*1}=\frac{-4}{2}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3}{2*1}=\frac{2}{2}
=1
$