4x-x^2=2+x

Solution for 4x-x^2=2+x equation:

4x-x^2=2+x

We move all terms to the left:

4x-x^2-(2+x)=0

We add all the numbers together, and all the variables

-x^2+4x-(x+2)=0

We add all the numbers together, and all the variables

-1x^2+4x-(x+2)=0

We get rid of parentheses

-1x^2+4x-x-2=0

We add all the numbers together, and all the variables

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

a = -1; b = 3; c = -2;
Δ = b2-4ac
Δ = 32-4·(-1)·(-2)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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